Let $$ABC$$  be a triangle in which the given vectors are represented by the sides $$AB$$   and $$AC$$
$$\eqalign{ & {\text{i}}{\text{.e}}{\text{., }}AB = \sqrt 3 \left( {a \times b} \right) \cr & {\text{and }}AC = b - \left( {a.b} \right)a \cr & \therefore \,AB.AC = \sqrt {3\left( {a \times b} \right)\left[ {b - \left( {a.b} \right)a} \right]} \cr & = \sqrt {3\left[ {\left( {a \times b} \right).b\left( {a.b} \right)\left( {a \times b} \right)a} \right]} \cr & = \sqrt {3\left( {0 - 0} \right)} \cr & = 0 \cr & {\text{Therefore, }}\angle BAC = {90^ \circ } \cr & A{B^2} = {\left| {\sqrt 3 \left( {a \times b} \right)} \right|^2} = 3{\left( {a \times b} \right)^2} \cr & A{C^2} = {\left[ {b - \left( {a.b} \right)a} \right]^2}......\left( {\text{i}} \right) \cr & = {\left( b \right)^2} + {\left( {a.b} \right)^2}{a^2} - 2\left( {b.a} \right)\left( {a.b} \right) \cr & = {\left( b \right)^2} + {\left( {a.b} \right)^2} - 2{\left( {a.b} \right)^2} \cr & = {\left( b \right)^2} - {\left( {a.b} \right)^2} = {\left( b \right)^2} = {\left| a \right|^2}{\left| b \right|^2}{\cos ^2}\theta \cr & = {\left( b \right)^2}\left[ {1 - {{\left| a \right|}^2}{{\cos }^2}\theta } \right] \cr & = {\left( b \right)^2}\left( {1 - {{\cos }^2}\theta } \right) \cr & = {\left( b \right)^2}{\sin ^2}\theta \cr & = {\left| a \right|^2}{\left| b \right|^2}{\sin ^2}\theta \cr & = {\left( {a \times b} \right)^2}......\left( {{\text{ii}}} \right) \cr & {\text{Dividing equation }}\left( {\text{i}} \right){\text{ by }}\left( {{\text{ii}}} \right){\text{ we get}} \cr & \frac{{A{B^2}}}{{A{C^2}}} = \frac{{3{{\left( {a \times b} \right)}^2}}}{{{{\left( {a \times b} \right)}^2}}} \cr & \Rightarrow A{B^2} = 3.A{C^2} \cr & \Rightarrow AB = \sqrt 3 AC \cr & \tan \,C = \frac{{AB}}{{AC}} = \frac{{\sqrt 3 AC}}{{AC}} = \sqrt 3 \cr & \therefore \,\angle C = {60^ \circ } \cr & \therefore \,\angle A = {180^ \circ } - {90^ \circ } - {60^ \circ } = {30^ \circ } \cr & {\text{Hence, angles of the triangle are 3}}{0^ \circ },\,{90^ \circ }{\text{ and }}{60^ \circ } \cr} $$

$$\eqalign{ & \overrightarrow {OA} = \overrightarrow i - \overrightarrow j + 2\overrightarrow k \,\,\,\,\,\,\,\,\,\,\overrightarrow {OB} = 2\overrightarrow i + \overrightarrow j + \overrightarrow k \cr & \therefore \overrightarrow {OC} = \overrightarrow {AB} = \overrightarrow {OB} - \overrightarrow {OA} = \overrightarrow i + 2\overrightarrow j - \overrightarrow k \cr & \therefore \,\overrightarrow {BC} = \overrightarrow {OC} - \overrightarrow {OB} = \left( {\overrightarrow i + 2\overrightarrow j - \overrightarrow k } \right) - \left( {2\overrightarrow i + \overrightarrow j + \overrightarrow k } \right) \cr & = - \overrightarrow i + \overrightarrow j - 2\overrightarrow k \cr} $$

We observe that
$$\eqalign{ & \overrightarrow a .\overrightarrow {{b_1}} = \overrightarrow a .\overrightarrow b - \left( {\frac{{\overrightarrow {b.} \overrightarrow a }}{{{{\left| {\overrightarrow a } \right|}^2}}}} \right)\overrightarrow a .\overrightarrow a = \overrightarrow a .\overrightarrow b - \overrightarrow a .\overrightarrow b = 0 \cr & \overrightarrow a .\overrightarrow {{c_2}} = \overrightarrow a \left( {\overrightarrow c - \frac{{\overrightarrow c .\overrightarrow a }}{{{{\left| {\overrightarrow a } \right|}^2}}}\overrightarrow a - \frac{{\overrightarrow c .\overrightarrow {{b_1}} }}{{{{\left| {\overrightarrow {{b_1}} } \right|}^2}}}\overrightarrow {{b_1}} } \right) \cr & = \overrightarrow a .\overrightarrow c - \overrightarrow c .\frac{{\overrightarrow a .\overrightarrow c }}{{{{\left| {\overrightarrow a } \right|}^2}}}{\left| {\overrightarrow a } \right|^2} - \frac{{\overrightarrow c .\overrightarrow {{b_1}} }}{{{{\left| {\overrightarrow {{b_1}} } \right|}^2}}}\left( {\overrightarrow a .\overrightarrow {{b_1}} } \right) \cr & = \overrightarrow a .\overrightarrow c - \overrightarrow a .\overrightarrow c - 0\,\,\,\,\,\,\,\,\,\left[ {\because \,\,\overrightarrow a .\overrightarrow {{b_1}} = 0} \right] \cr & = 0 \cr & {\text{And }}\overrightarrow {{b_1}} .\overrightarrow {{c_2}} = \overrightarrow {{b_1}} .\left( {\overrightarrow c - \frac{{\overrightarrow c .\overrightarrow a }}{{{{\left| {\overrightarrow a } \right|}^2}}}\overrightarrow a - \frac{{\overrightarrow c .\overrightarrow {{b_1}} }}{{{{\left| {\overrightarrow {{b_1}} } \right|}^2}}}\overrightarrow {{b_1}} } \right) \cr & = \overrightarrow {{b_1}} .\overrightarrow c - \frac{{\left( {\overrightarrow c .\overrightarrow a } \right)\left( {\overrightarrow {{b_1}} .\overrightarrow a } \right)}}{{{{\left| {\overrightarrow a } \right|}^2}}} - \frac{{\overrightarrow c .\overrightarrow {{b_1}} }}{{{{\left| {\overrightarrow {{b_1}} } \right|}^2}}}\overrightarrow {{b_1}} .\overrightarrow {{b_1}} \cr & = \overrightarrow {{b_1}} .\overrightarrow c - 0 - \overrightarrow {{b_1}} .\overrightarrow c \,\,\,\,\,\,\,\,\,\left[ {{\text{Using }}\overrightarrow {{b_1}} .\overrightarrow a = 0} \right] \cr & = 0 \cr} $$
Hence $$\overrightarrow a .\overrightarrow {{b_1}} = \overrightarrow a .\overrightarrow {{c_2}} = \overrightarrow {{b_1}} .\overrightarrow {{c_2}} = 0$$
$$ \Rightarrow \left( {\overrightarrow a ,\overrightarrow {{b_1}} ,\,\overrightarrow {{c_2}} } \right)$$     is a set of orthogonal vectors.

$$\overrightarrow a .\left( {\overrightarrow b + \overrightarrow c } \right) = 0{\text{ or }}\overrightarrow a .\overrightarrow b + \overrightarrow a .\overrightarrow c = 0$$
Similarly, we get $$\overrightarrow b .\overrightarrow c + \overrightarrow b .\overrightarrow a = 0$$     and $$\overrightarrow c .\overrightarrow a + \overrightarrow c .\overrightarrow b = 0$$
Adding these, $$\overrightarrow a .\overrightarrow b + \overrightarrow b .\overrightarrow c
+ \overrightarrow c .\overrightarrow a = 0$$
$$\eqalign{ & {\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right|^2} = {\left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right)^2} \cr & = {\left| {\overrightarrow a } \right|^2} + {\left| {\overrightarrow b } \right|^2} + {\left| {\overrightarrow c } \right|^2} + 2\left( {\overrightarrow a .\overrightarrow b + \overrightarrow b .\overrightarrow c + \overrightarrow c .\overrightarrow a } \right) \cr & = 1 + 2 + 3 + 0 \cr & = 6 \cr & \therefore \,\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right| = \sqrt 6 \cr} $$

Let the required vector be $$\hat i + \hat j$$
Since the vector $$\hat i + \hat j$$  is equally inclined to the vectors $$\hat i + 3\hat j$$  and $$3\hat i + \hat j$$  therefore
Angle between $$\hat i + \hat j$$  and $$\hat i + 3\hat j = {\theta _1}$$   is equal to angle between $$\hat i + \hat j$$  and $$3\hat i + \hat j = {\theta _2}$$
$$\therefore $$  Angle between $$\hat i + \hat j$$  and $$\hat i + 3\hat j$$
$$\eqalign{ & = {\cos ^{ - 1}}\left[ {\frac{{\left( 1 \right)\left( 1 \right) + \left( 1 \right)\left( 3 \right)}}{{\sqrt {{{\left( 1 \right)}^2} + {{\left( 1 \right)}^2}} \sqrt {{{\left( 1 \right)}^2} + {{\left( 3 \right)}^2}} }}} \right] \cr & = {\cos ^{ - 1}}\left[ {\frac{{1 + 3}}{{\sqrt 2 \sqrt {10} }}} \right] \cr & = {\cos ^{ - 1}}\left[ {\frac{4}{{\sqrt 2 \sqrt {10} }}} \right] \cr & = {\cos ^{ - 1}}\left[ {\frac{2}{{\sqrt 5 }}} \right] \cr & {\text{and angle between }}\hat i + \hat j\,{\text{and}}\,3\hat i + \hat j \cr & = {\cos ^{ - 1}}\left| {\frac{{1 + 3}}{{\sqrt {10} \sqrt 2 }}} \right| \cr & = {\cos ^{ - 1}}\left( {\frac{4}{{\sqrt 2 \sqrt {10} }}} \right) \cr & = {\cos ^{ - 1}}\left( {\frac{2}{{\sqrt 5 }}} \right) \cr & {\text{Hence, required vector is }}\hat i + \hat j \cr} $$

$$\eqalign{ & {\text{Here, }}\overrightarrow a .\left( {\overrightarrow a + \overrightarrow b } \right) = 0{\text{ or }}{\left| {\overrightarrow a } \right|^2} + \overrightarrow a .\overrightarrow b = 0\,\,\,{\text{or }}\frac{1}{2}{\text{ }}{\left| {\overrightarrow b } \right|^2} + \overrightarrow a .\overrightarrow b = 0 \cr & \left( {2\overrightarrow a + \overrightarrow b } \right).\overrightarrow b = 2\overrightarrow a .\overrightarrow b + {\left| {\overrightarrow b } \right|^2} = 2\left\{ {\overrightarrow a .\overrightarrow b + \frac{1}{2}{{\left| {\overrightarrow b } \right|}^2}} \right\} = 0 \cr & \therefore \,\left( {2\overrightarrow a + \overrightarrow b } \right) \bot \overrightarrow b \cr & \left( {2\overrightarrow a - \overrightarrow b } \right).\overrightarrow b = \,2\overrightarrow a .\overrightarrow b - {\left| {\overrightarrow b } \right|^2} = 2\overrightarrow a .\overrightarrow b + 2\overrightarrow a .\overrightarrow b = 4\overrightarrow a .\overrightarrow b \ne 0 \cr & \overrightarrow a .\left( {2\overrightarrow a + \overrightarrow b } \right) = 2{\left| {\overrightarrow a } \right|^2} + \overrightarrow a .\overrightarrow b = {\left| {\overrightarrow a } \right|^2} + \left( {{{\left| {\overrightarrow a } \right|}^2} + \overrightarrow a .\overrightarrow b } \right) = {\left| {\overrightarrow a } \right|^2} \ne 0 \cr} $$

The points are not on the same plane.

$$\eqalign{ & {\text{We have, }}\left[ {\vec a \times \,\vec b\,\,\vec b \times \,\vec c\,\,\vec c \times \,\vec a} \right] \cr & = \left( {\vec a \times \,\vec b} \right).\left\{ {\left( {\vec b \times \,\vec c} \right) \times \left( {\vec c \times \,\vec a} \right)} \right\} \cr & = \left( {\vec a \times \,\vec b} \right).\left\{ {\left( {\vec m.\vec a} \right)\vec c - \left( {\vec m.\vec c} \right)\vec a} \right\}\,\,\,\,\,\,\left( {{\text{where }}\vec m = \vec b \times \,\vec c} \right) \cr & = \left\{ {\left( {\vec a \times \,\vec b} \right).\vec c} \right\}.\left\{ {\vec a.\left( {\vec b \times \,\vec c} \right)} \right\} \cr & = {\left[ {\vec a,\,\vec b,\,\vec c} \right]^2} = {4^2} = 16 \cr} $$

$$\eqalign{ & \overrightarrow a .\overrightarrow b = \left( {2{x^2}\hat i + 4x\hat j + \hat k} \right).\left( {7\hat i - 2\hat j + x\hat k} \right) \cr & \Rightarrow \overrightarrow a .\overrightarrow b = 2{x^2}\left( 7 \right) + \left( {4x} \right)\left( { - 2} \right) + \left( 1 \right)\left( x \right) \cr & \Rightarrow \overrightarrow a .\overrightarrow b = 14{x^2} - 7x \cr & \hat i.\hat i = 1s = \hat j.\hat j = 1 \cr} $$
The angle between vectors $$\overrightarrow a $$ and $$\overrightarrow b $$ is obtuse
$$\eqalign{ & \Rightarrow \overrightarrow a .\overrightarrow b < 0 \cr & \Rightarrow 14{x^2} - 7x < 0 \cr & \Rightarrow 7x\left( {2x - 1} \right) < 0 \cr & \Rightarrow 14x\left( {x - \frac{1}{2}} \right) < 0 \cr} $$
$$ \Rightarrow x$$  lies between $$0$$ and $$\frac{1}{2}$$ (By the Method of Intervals) i.e., $$0 < x < \frac{1}{2}$$
Hence, the angle between the given vectors is obtuse if $$x\, \in \left( {0,\,\frac{1}{2}} \right).$$

$$\overrightarrow p \times \overrightarrow q $$   is perpendicular to $$\overrightarrow p $$ and $$\overrightarrow q .$$ So $$\overrightarrow p ,\,\overrightarrow q ,\,\overrightarrow p \times \overrightarrow q $$   are noncoplanar. Therefore, the given linear relation is possible only when $$b - c = 0,\,c - a = 0,\,a - b = 0.$$       So, $$a = b = c.$$