$$\eqalign{ & {\text{Here,}} \cr & \frac{{\left( {3 - \left( { - 2} \right)} \right)}}{{1 - 3}} = \frac{{ - 6 - 4}}{{ - 2 - \left( { - 6} \right)}} = \frac{{ - 8 - 7}}{{ - 2 - \left( { - 8} \right)}} \cr & \Rightarrow - \frac{5}{2} = - \frac{5}{2} = - \frac{5}{2} \cr & {\text{Obviously, points are collinear}}{\text{.}} \cr} $$

Equation of plane through $$\left( {1,\,0,\,0} \right)$$  is
$$a\left( {x - 1} \right) + by + cz = 0.....({\text{i}})$$
(i) passes through $$\left( {0,\,1,\,0} \right)$$
$$\eqalign{ & - a + b = 0\,\,\, \Rightarrow b = a\,; \cr & {\text{Also, }}\,\cos \,{45^ \circ } = \frac{{a + a}}{{\sqrt {2\left( {2{a^2} + {c^2}} \right)} }} \cr & \Rightarrow 2a = \sqrt {2{a^2} + {c^2}} \cr & \Rightarrow 2{a^2} = {c^2} \cr & \Rightarrow c = \sqrt 2 a \cr} $$
So d.r of normal area, $$a,\,a,\,\sqrt 2 a\,\,{\text{i}}{\text{.e}}{\text{., }}1,\,1,\,\sqrt 2 $$

We have, $$z = 0$$  for the point where the line intersects the curve.
Therefore,
$$\eqalign{ & \frac{{x - 2}}{3} = \frac{{y + 1}}{2} = \frac{{0 - 1}}{{ - 1}} \cr & \Rightarrow \frac{{x - 2}}{3} = 1{\text{ and }}\frac{{y + 1}}{2} = 1 \cr & \Rightarrow x = 5{\text{ and }}y = 1 \cr} $$
Put these value in $$xy = {c^2},$$   we get,
$$5 = {c^2} \Rightarrow c = \pm 5$$

Equation of the line through $$\left( {1,\, - 2,\,3} \right)$$   parallel to the line $$\frac{x}{2} = \frac{y}{3} = \frac{{z - 1}}{{ - 6}}$$     is
$$\frac{{x - 1}}{2} = \frac{{y + 2}}{3} = \frac{{z - 3}}{{ - 6}} = r\,\,\,\,\left( {{\text{say}}} \right)......\left( 1 \right)$$
Then any point on $$\left( 1 \right)$$ is $$\left( {2r + 1,\,3r - 2,\, - 6r + 3} \right)$$
If this point lies on the plane $$x - y + z = 5$$    then
$$\left( {2r + 1} \right) - \left( {3r - 2} \right) + \left( { - 6r + 3} \right) = 5 \Rightarrow r = \frac{1}{7}$$
Hence the point is $$\left( {\frac{9}{7},\, - \frac{{11}}{7},\,\frac{{15}}{7}} \right)$$
Distance between $$\left( {1,\, - 2,\,3} \right)$$   and $$\left( {\frac{9}{7},\, - \frac{{11}}{7},\,\frac{{15}}{7}} \right)$$
$$\eqalign{ & = \sqrt {\left( {\frac{4}{{49}} + \frac{9}{{49}} + \frac{{36}}{{49}}} \right)} \cr & = \sqrt {\left( {\frac{{49}}{{49}}} \right)} \cr & = 1 \cr} $$

Equation of the plane containing the lines
$$\eqalign{ & 2x - 5y + z = 3{\text{ and }}x + y + 4z = 5{\text{ is}} \cr & 2x - 5y + z - 3 + \lambda \left( {x + y + 4z - 5} \right){\text{ = 0}} \cr & \Rightarrow \left( {2 + \lambda } \right)x + \left( { - 5 + \lambda } \right)y + \left( {1 + 4\lambda } \right)z + \left( { - 3 - 5\lambda } \right) = 0.....({\text{i}}) \cr} $$
Since the plane (i) parallel to the given plane $$x+3y+6z=1$$
$$\therefore \frac{{2 + \lambda }}{1} = \frac{{ - 5 + \lambda }}{3} = \frac{{1 + 4\lambda }}{6}\,\,\,\,\, \Rightarrow \lambda = - \frac{{11}}{2}$$
Hence equation of the required plane is
$$\eqalign{ & \left( {2 - \frac{{11}}{2}} \right)x + \left( { - 5 - \frac{{11}}{2}} \right)y + \left( {1 - \frac{{44}}{2}} \right)z + \left( { - 3 + \frac{{55}}{2}} \right) = 0 \cr & \Rightarrow x + 3y + 6z = 7 \cr} $$

Given equation of a plane is $$x-2y+2z-5=0$$
So, Equation of parallel plane is given by $$x-2y+2z+d=0$$
Now, it is given that distance from origin to the parallel plane is 1.
$$\therefore \left| {\frac{d}{{\sqrt {{1^2} + {2^2} + {2^2}} }}} \right| = 1\,\,\,\,\,\, \Rightarrow d = \pm 3$$
So equation of required plane $$x - 2y + 2z \pm 3 = 0$$

$$\left( {\bf{A}} \right)$$   Here $$A\left( {0,\, - 1} \right),\,B\left( {2,\,1} \right),\,C\left( {0,\,3} \right),\,D\left( { - 2,\,1} \right)$$
For a rhombus all four sides are equal but the diagonal are not equal, we see
$$AC = \sqrt {0 + {4^2}} = 4,\,\,BD = \sqrt {{4^2} - 0} = 4$$
Since diagonals are equals therefore it is a square, not rhombus.
$$\left( {\bf{B}} \right)$$   Here $$AB = \sqrt {{2^2} + {{\left( { - 3} \right)}^2}} = \sqrt {13} ,\,\,BC = \sqrt {{6^2} + {4^2}} = \sqrt {52} $$
Since $$AB \ne BC$$   therefore it is not square.
$$\left( {\bf{C}} \right)$$   In this case mid point of $$AC$$  is $$\left( {\frac{{4 - 2}}{2},\,\frac{{3 - 1}}{2}} \right){\text{ or }}\left( {1,\,1} \right)$$
Also mid-point of diagonal $$BD\left( {\frac{{1 + 1}}{2},\,\frac{{0 + 2}}{2}} \right){\text{ or }}\left( {1,\,1} \right)$$
Hence the points are vertices of a parallelogram.

Since line is parallel to the plane, vector $$2\overrightarrow i + 3\overrightarrow j + \lambda \overrightarrow k $$     is perpendicular to the normal to the plane $${2\overrightarrow i + 3\overrightarrow j + 4\overrightarrow k }$$
$$ \Rightarrow 2 \times 2 + 3 \times 3 + 4\lambda = 0\,{\text{or }}\lambda = - \frac{{13}}{4}$$

We know, if $${a_1}x + {b_1}y + {c_1}z = {d_1}$$     and $${a_2}x + {b_2}y + {c_2}z = {d_2}$$
are two planes then angle between them is
$$\cos \,\theta = \frac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\sqrt {a_1^2 + b_1^2 + c_1^2} \,\sqrt {a_2^2 + b_2^2 + c_2^2} }}$$
Let $$q$$ be the angle between given planes. Here,
$$\eqalign{ & {a_1} = 2,\,{b_1} = - 1,\,{c_1} = 1\,;\,{a_2} = 1,\,{b_2} = 1,\,{c_2} = 2 \cr & \therefore \,\cos \,q = \frac{{2 \times 1 + 1 \times \left( { - 1} \right) + 1 \times 2}}{{\sqrt {4 + 1 + 1} \,\sqrt {1 + 1 + 4} }} \cr & \Rightarrow \cos \,q = \frac{3}{6} = \frac{1}{2} \cr & \Rightarrow \cos \,q = \cos \,\frac{\pi }{3} \cr & \Rightarrow \theta = \frac{\pi }{3} \cr} $$

The given equation of the planes are $$x + y + 2z = 3$$    and $$ - 2x + y - z = 11$$
We know that, the angle between the planes
$$\eqalign{ & {a_1}x + {b_1}y + {c_1}z + {d_1} = 0{\text{ and}} \cr & {a_2}x + {b_2}y + {c_2}z + {d_2} = 0{\text{ is given by}} \cr & \cos \,\theta = \left| {\frac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\sqrt {a_1^2 + b_1^2 + c_1^2} \,\sqrt {a_2^2 + b_2^2 + c_2^2} }}} \right| \cr & {\text{Here, }}{a_1} = 1,\,{b_1} = 1,\,{c_1} = 2,\,{a_2} = - 2,\,{b_2} = 1,\,{c_2} = - 1 \cr & \therefore \,\cos \,\theta = \left| {\frac{{1 \times \left( { - 2} \right) + 1 \times 1 + 2 \times \left( { - 1} \right)}}{{\sqrt {1 + 1 + 4} \,\sqrt {4 + 1 + 1} }}} \right| \cr & \Rightarrow \cos \,\theta = \left| {\frac{{ - 2 + 1 - 2}}{{\sqrt 6 \,\sqrt 6 }}} \right| \cr & \Rightarrow \cos \,\theta = \left| {\frac{3}{6}} \right| \cr & \Rightarrow \cos \,\theta = \frac{1}{2} \cr & \Rightarrow \cos \,\theta = \cos \frac{\pi }{3} \cr & \Rightarrow \theta = \frac{\pi }{3}\, \cr} $$