A point on line is $$\left( {2,\, - 2,\,3} \right)$$   its perpendicular distance from the plane $$x+5y+z-5=0$$     is
$$ = \left| {\frac{{2 - 10 + 3 - 5}}{{\sqrt {1 + 25 + 1} }}} \right| = \frac{{10}}{{3\sqrt 3 }}$$

If the given points $$\left( {\lambda ,\,\mu ,\, - 6} \right),\,\left( {3,\,2,\, - 4} \right)$$     and $$\left( {9,\,8,\, - 10} \right)$$   are collinear then
$$\eqalign{ & \frac{{\lambda - 3}}{{9 - 3}} = \frac{{\mu - 2}}{{8 - 2}} = \frac{{ - 6 + 4}}{{ - 10 + 4}} \cr & \Rightarrow \lambda = 5,\,\mu = 4 \cr} $$

Given equation of planes are :
$$x + y + z = 7$$    and $$\alpha x + \beta y + \gamma z = 3$$
For these planes to be parallel, coefficients of $$x,\,y$$  and $$z$$ should be same i.e.
$$ \Rightarrow \alpha = \beta = \gamma $$

Let the angle of line makes with the positive direction of $$z$$-axis is $$\alpha $$ direction cosines of line with the $$+ve$$  directions of $$x$$-axis, $$y$$-axis, and $$z$$-axis is $$1,\,m,\,n$$   respectively.
$$\therefore l = \cos \frac{\pi }{4},\,\,m = \cos \frac{\pi }{4},\,\,n = \cos \,\alpha $$
As we know that, $${l^2} + {m^2} + {n^2} = 1$$
$$\eqalign{ & \therefore {\cos ^2}\frac{\pi }{4} + {\cos ^2}\frac{\pi }{4} + {\cos ^2}\,\alpha = 1 \cr & \Rightarrow \frac{1}{2} + \frac{1}{2} + {\cos ^2}\,\alpha = 1 \cr & \Rightarrow {\cos ^2}\,\alpha = 0 \cr & \Rightarrow \alpha = \frac{\pi }{2} \cr} $$
Hence, angle with positive direction of the $$z$$-axis is $$\frac{\pi }{2}$$

Let the variable point be $$\left( {\alpha ,\,\beta ,\,\gamma } \right)$$   then according to question
$$\eqalign{ & {\left( {\frac{{\left| {\alpha + \beta + \gamma } \right|}}{{\sqrt 3 }}} \right)^2} + {\left( {\frac{{\left| {\alpha - \gamma } \right|}}{{\sqrt 2 }}} \right)^2} + {\left( {\frac{{\left| {\alpha - 2\beta + \gamma } \right|}}{{\sqrt 6 }}} \right)^2} = 9 \cr & \Rightarrow {\alpha ^2} + {\beta ^2} + {\gamma ^2} = 9 \cr} $$
So, the locus of the point is $${x^2} + {y^2} + {z^2} = 9.$$

For given sphere centre is $$\left( {3,\,6,\,1} \right)$$
Coordinates of one end of diameter of the sphere are $$\left( {2,\,3,\,5} \right).$$
Let the coordinates of the other end of diameter are $$\left( {\alpha ,\,\beta ,\,\gamma } \right)$$
$$\eqalign{ & \therefore \,\frac{{\alpha + 2}}{2} = 3,\,\,\frac{{\beta + 3}}{2} = 6,\,\,\frac{{\gamma + 5}}{2} = 1 \cr & \Rightarrow \alpha = 4,\,\beta = 9{\text{ and }}\,\gamma = - 3 \cr} $$
$$\therefore $$ Coordinate of other end of diameter are $$\left( {4,\,9,\, - 3} \right)$$

Let $$L$$ be the foot of perpendicular drawn from the point $$P\left( {6,\,7,\,8} \right)$$   to the $$xy$$ -plane and the distance of this foot $$L$$ from $$P$$ is $$z$$-coordinate of $$P$$, i.e., $$8$$ units.

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

Given lines in vector form are $$\vec r = \left( {\hat i - \hat j + \hat k} \right) + \lambda \left( {2\hat i + 3\hat j + 4\hat k} \right)$$
and $$\vec r = \left( {3\hat i + k\hat j} \right) + \mu \left( {\hat i + 2\hat j + \hat k} \right)$$
These will intersect if shortest distance between them $$=0$$
\[\begin{array}{l} {\rm{i}}{\rm{.e}}{\rm{.,}}\,\,\,\,\left( {{{\vec a}_2} - {{\vec a}_1}} \right).{{\vec b}_1} \times {{\vec b}_2} = 0\\ \Rightarrow \left| \begin{array}{l} 3 - 1\,\,\,\,\,k + 1\,\,\,\,\, - 1\\ \,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4\\ \,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 \end{array} \right| = 0\\ \Rightarrow 2\left( { - 5} \right) - \left( {k + 1} \right)\left( { - 2} \right) - 1\left( 1 \right) = 0\\ \Rightarrow k = \frac{9}{2} \end{array}\]

Let any point on the intersecting line
$$\frac{{x + 1}}{{ - 3}} = \frac{{y - 3}}{2} = \frac{{z - 2}}{{ - 1}} = \lambda \,\,\,\,\left( {{\text{say}}} \right)$$
is $$\left( { - 3\lambda - 1,\,2\lambda + 3,\, - \lambda + 2} \right)$$
Since, the above point lies on a line which passes through the point $$\left( { - 4,\,3,\,1} \right)$$
Then, direction ratio of the required line
$$\eqalign{ & = \left\langle { - 3\lambda - 1 + 4,\,2\lambda + 3 - 3,\, - \lambda + 2 - 1} \right\rangle \cr & {\text{or,}}\,\left\langle { - 3\lambda + 3,\,2\lambda ,\, - \lambda + 1} \right\rangle \cr} $$
Since, line is parallel to the plane $$x+2y-z-5=0$$
Then, perpendicular vector to the line is $$\hat i + 2\hat j - \hat k$$
$$\eqalign{ & {\text{Now }}\left( { - 3\lambda + 3} \right)\left( a \right) + \left( {2\lambda } \right)\left( b \right) + \left( { - \lambda + 1} \right)\left( { - 1} \right) = 0 \cr & \Rightarrow \lambda = - 1 \cr} $$
Now direction ratio of the required line
$$ = \left\langle {6,\, - 2,\,2} \right\rangle {\text{ or }}\left\langle {3,\, - 1,\,1} \right\rangle $$
Hence required equation of the line is $$\frac{{x + 4}}{3} = \frac{{y - 3}}{{ - 1}} = \frac{{z - 1}}{1}$$