For $$\left( {\frac{1}{{\sqrt 2 }},\,\frac{1}{2},\,k} \right)$$   to represent direction cosines, we should have
$$\eqalign{ & {\left( {\frac{1}{{\sqrt 2 }}} \right)^2} + {\left( {\frac{1}{2}} \right)^2} + {k^2} = 1\,\,{\text{or, }}\frac{1}{2} + \frac{1}{4} + {k^2} + 1 \cr & \Rightarrow k = \pm \frac{1}{2} \cr} $$

$$\because $$ The line $$\frac{{x - 2}}{3} = \frac{{y - 1}}{{ - 5}} = \frac{{z + 2}}{2}$$     lies in the plane $$x + 3y - \alpha z + \beta = 0$$
$$\therefore Pt\left( {2,\,1,\, - 2} \right)$$    lies on the plane
i.e., $$2 + 3 + 2\alpha + \beta = 0$$
or $$2\alpha + \beta + 5 = 0.....({\text{i}})$$
Also normal to plane will be perpendicular to line,
$$\eqalign{ & \therefore 3 \times 1 - 5 \times 3 + 2 \times \left( { - \alpha } \right) = 0 \cr & \Rightarrow \alpha = - 6 \cr} $$
From equation (i) then, $$\beta = 7$$
$$\therefore \,\,\left( {\alpha ,\,\beta } \right) = \left( { - 6,\,7} \right)$$

Centers of given spheres are $$\left( { - 3,\,4,\,1} \right)$$   and $$\left( {5,\, - 2,\,1} \right).$$
Mid point of centres is (1, 1, 1).
Satisfying this in the equation of plane, we get
$$\eqalign{ & 2a - 3a + 4a + 6 = 0 \cr & \Rightarrow a = - 2 \cr} $$

$$\eqalign{ & {\text{Since d}}{\text{.c}}{\text{. of line are }}\left\{ {\frac{1}{c},\,\frac{1}{c},\,\frac{1}{c}} \right\} \cr & \therefore \,\frac{1}{{{c^2}}} + \frac{1}{{{c^2}}} + \frac{1}{{{c^2}}} = 1 \cr & \Rightarrow {c^2} = 3 \cr & \Rightarrow c = \pm \sqrt 3 \cr} $$

Let us take a triangle $$ABC$$  and their vertices $$A\left( {a,\,0,\,0} \right),\,B\left( {0,\,b,\,0} \right)$$     and $$C\left( {0,\,0,\,c} \right)$$
Therefore the equation of plane is
$$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1......\left( {\text{i}} \right)$$
Now, given centroid of $$\Delta ABC$$   is $$\left( {\alpha ,\,\beta ,\,\gamma } \right)$$
As we know, centroid of $$\Delta ABC$$   with vertices $$\left( {{x_1},\,{y_1},\,{z_1}} \right),\,\left( {{x_2},\,{y_2},\,{z_2}} \right)$$       and $$\left( {{x_3},\,{y_3},\,{z_3}} \right)$$   is given by
$$\left( {\frac{{{x_1} + {x_2} + {x_3}}}{3},\,\frac{{{y_1} + {y_2} + {y_3}}}{3},\,\frac{{{z_1} + {z_2} + {z_3}}}{3}} \right)$$
$$\therefore $$  By using this formula, we have
$$\eqalign{ & \frac{{a + 0 + 0}}{3} = \alpha \, \Rightarrow a = 3\alpha \,; \cr & \frac{{0 + b + 0}}{3} = \beta \, \Rightarrow b = 3\beta \,; \cr & {\text{and }}\frac{{0 + 0 + c}}{3} = \gamma \, \Rightarrow c = 3\gamma \cr} $$
Now, put the values of $$a,\,b,\,c$$  in equation $$\left( {\text{i}} \right)$$, which gives
$$\eqalign{ & \frac{x}{{3\alpha }} + \frac{y}{{3\beta }} + \frac{z}{{3\gamma }} = 1 \cr & \therefore \,\frac{x}{\alpha } + \frac{y}{\beta } + \frac{z}{\gamma } = 3 \cr} $$

$$\because $$  DR's of the normal of the plane $$ = 2, - 3, 6$$
and DR's of $$x$$-axis is $$1,\,0,\,0$$
and If $$\theta $$ is the angle between the plane and $$x$$-axis, then
$$\eqalign{ & \sin \,\theta = \frac{{2\left( 1 \right) + \left( { - 3} \right)\left( 0 \right) + \left( 6 \right)\left( 0 \right)}}{{\sqrt {{2^2} + {{\left( { - 3} \right)}^2} + {6^2}} \,\sqrt 1 }} \cr & \Rightarrow \sin \,\theta = \frac{2}{7} \cr & \Rightarrow \theta = {\sin ^{ - 1}}\left( {\frac{2}{7}} \right) \cr & \therefore \,a = \frac{2}{7} \cr} $$

Given equations of two planes are $$2x - y + z = 4$$    and $$x + y + 2z = 6$$
So, angle between them is
$$\eqalign{ & \cos \,\theta = \frac{{2\left( 1 \right) + \left( { - 1} \right)\left( 1 \right) + \left( 1 \right)\left( 2 \right)}}{{\sqrt {4 + 1 + 1} \,\sqrt {4 + 1 + 1} }} \cr & \Rightarrow \cos \,\theta = \frac{{2 - 1 + 2}}{{\sqrt 6 \,\sqrt 6 }} \cr & \Rightarrow \cos \,\theta = \frac{3}{6} \cr & \Rightarrow \cos \,\theta = \frac{1}{2} \cr & \Rightarrow \theta = \frac{\pi }{3} \cr} $$

Let the point $$A\left( {x,\,y,\,z} \right)$$   is equidistant from the points $$B\left( {a,\,0,\,0} \right),\,C\left( {0,\,a,\,0} \right),\,D\left( {0,\,0,\,a} \right)$$       and $$E\left( {0,\,0,\,0} \right).$$
$$\eqalign{ & {\text{Hence,}} \cr & \Rightarrow {\left( {x - a} \right)^2} + {y^2} + {z^2} = {x^2} + {\left( {y - a} \right)^2} + {z^2} \cr & \Rightarrow {\left( {x - a} \right)^2} + {y^2} + {z^2} = {x^2} + {y^2} + {\left( {z - a} \right)^2} \cr & \Rightarrow {\left( {x - a} \right)^2} + {y^2} + {z^2} = {x^2} + {y^2} + {z^2} \cr & \Rightarrow {\left( {x - a} \right)^2} + {y^2} + {z^2} = {x^2} + {\left( {y - a} \right)^2} + {z^2} \cr & \Rightarrow {x^2} + {a^2} - 2ax + {y^2} + {z^2} = {x^2} + {y^2} + {a^2} - 2ay + {z^2} \cr & \Rightarrow - 2ax = - 2ay \cr & \Rightarrow ax = ay \cr & \Rightarrow x = y \cr & {\text{Similarly,}} \cr & ay = az \cr & \Rightarrow y = z \cr & \Rightarrow x = y = z \cr & \therefore \,{\left( {x - a} \right)^2} + {x^2} + {x^2} = {x^2} + {x^2} + {x^2} \cr & \Rightarrow {x^2} + {a^2} - 2ax + {x^2} + {x^2} = 3{x^2} \cr & \Rightarrow {a^2} = 2ax \cr & \Rightarrow x = \frac{a}{2} \cr & \therefore \,{\text{Point is }}\left( {\frac{a}{2},\,\frac{a}{2},\,\frac{a}{2}} \right) \cr} $$

Suppose $$xy$$ -plane divides the line joining the given points in the ratio $$\lambda :1.$$
The coordinate of the point of division are
$$\left( {\frac{{2\lambda - 1}}{{\lambda + 1}},\,\frac{{ - 5\lambda + 3}}{{\lambda + 1}},\,\frac{{6\lambda + 4}}{{\lambda + 1}}} \right)$$
This point lies on $$xy$$ -plane
$$\frac{{6\lambda + 4}}{{\lambda + 1}} = 0 \Rightarrow \lambda = - \frac{3}{2}$$
Hence, $$xy$$ -plane divides externally in the ratio $$3 : 2.$$

$$\eqalign{ & {\text{Let the point be }}P\left( {x,\,y,\,z} \right){\text{ and two points,}} \cr & \left( {a,\,0,\,0} \right){\text{ and }}\left( { - a,\,0,\,0} \right){\text{ be }}A{\text{ and }}B \cr & {\text{As given in the problem,}}\,P{A^2} + P{B^2} = 2{c^2} \cr & {\text{So,}} \cr & {\left( {x + a} \right)^2} + {\left( {y - 0} \right)^2} + {\left( {z - 0} \right)^2} + {\left( {x - a} \right)^2} + {\left( {y - 0} \right)^2} + {\left( {z - 0} \right)^2} = 2{c^2} \cr & {\text{or,}}\,{\left( {x + a} \right)^2} + {y^2} + {z^2} + {\left( {x - a} \right)^2} + {y^2} + {z^2} = 2{c^2} \cr & \Rightarrow {x^2} + 2a + {a^2} + {y^2} + {z^2} + {x^2} - 2a + {a^2} + {y^2} + {z^2} = 2{c^2} \cr & \Rightarrow 2\left( {{x^2} + {y^2} + {z^2} + {a^2}} \right) = 2{c^2} \cr & \Rightarrow {x^2} + {y^2} + {z^2} + {a^2} = {c^2} \cr & \Rightarrow {x^2} + {a^2} = {c^2} - {y^2} - {z^2} \cr} $$