Let $$A\left( {1,\, - 2,\,3} \right)$$   and $$B\left( {4,\,2,\, - 1} \right)$$
Let the plane $$XOY$$  meet the line $$AB$$  in the point $$C$$ such that $$C$$ divides $$AB$$  in the ratio $$k : 1,$$  then
$$C \equiv \left( {\frac{{4k + 1}}{{k + 1}},\,\frac{{2k - 2}}{{k + 1}},\,\frac{{ - k + 3}}{{k + 1}}} \right)$$
Since $$C$$ lies on the plane $$XOY$$  i.e., the plane $$z = 0,$$  therefore
$$\frac{{ - k + 3}}{{k + 1}} = 0 \Rightarrow k = 3$$

Obviously the line perpendicular to the plane because $$\frac{a}{a} = \frac{b}{b} = \frac{c}{c}$$   i.e., their direction ratios are proportional.

Centre of sphere $$ = \left( { - 1,\,1,\,2} \right)$$
Radius of sphere $$\sqrt {1 + 1 + 4 + 19} = 5$$
Perpendicular distance from centre to the plane
$$\eqalign{ & OC = d = \left| {\frac{{ - 1 + 2 + 4 + 7}}{{\sqrt {1 + 4 + 4} }}} \right| = \frac{{12}}{3} = 4 \cr & A{C^2} = A{O^2} - O{C^2} \cr & A{C^2} = {5^2} - {4^2} = 9 \cr & AC = 3 \cr} $$

$$\eqalign{ & {\text{Equation of the plane containing the lines}} \cr & 2x - 5y + z = 3{\text{ and }}\,x + y + 4y = 5{\text{ is}} \cr & 2x - 5y + z - 3 + \lambda \left( {x + y + 4y - 5} \right) = 0 \cr & \Rightarrow \left( {2 + \lambda } \right)x + \left( { - 5 + \lambda } \right)y + \left( {1 + 4\lambda } \right)z + \left( { - 3 - 5\lambda } \right) = 0......\left( {\text{i}} \right) \cr & {\text{Since the plane }}({\text{i}}){\text{ parallel to the given plane}} \cr & x + 3y + 6z = 1 \cr & \therefore \,\frac{{2 + \lambda }}{1} = \frac{{ - 5 + \lambda }}{3} = \frac{{1 + 4\lambda }}{6} \Rightarrow \lambda = - \frac{{11}}{2} \cr & {\text{Hence equation of the required plane is}} \cr & \left( {2 - \frac{{11}}{2}} \right)x + \left( { - 5 - \frac{{11}}{2}} \right)y + \left( {1 - \frac{{44}}{2}} \right)z + \left( { - 3 - \frac{{55}}{2}} \right) = 0 \cr & \Rightarrow \left( {4 - 11} \right)x + \left( { - 10 - 11} \right)y + \left( {2 - 44} \right)z + \left( { - 6 + 55} \right) = 0 \cr & \Rightarrow x + 3y + 6z = 7 \cr} $$

$$\eqalign{ & \cos \,2\alpha + \cos \,2\beta + \cos \,2\gamma + {\sin ^2}\alpha + {\sin ^2}\beta + {\sin ^2}\gamma \cr & = \cos {\,^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma \cr & = 1 \cr} $$

$$\eqalign{ & {\text{The given lines are : - }} \cr & \frac{{x - 2}}{1} = \frac{{y - \left( { - 1} \right)}}{{ - 2}} = \frac{{z - \left( { - 2} \right)}}{1}{\text{ and}} \cr & \frac{{x - 1}}{1} = \frac{{y - \left( { - \frac{3}{2}} \right)}}{{\frac{3}{2}}} = \frac{{z - \left( { - 5} \right)}}{2}\, \cr & {\text{dr's of Ist line are}}\,:{\text{ - }} \cr & {a_1} = 1,\,{b_1} = - 2,\,{c_1} = 1 \cr & {\text{dr's of IInd line are}}\,{\text{: - }} \cr & {a_2} = 2,\,{b_2} = 3,\,{c_2} = 4 \cr & {\text{Let }}'\theta '\,{\text{be the angle between two lines, then,}} \cr & \cos \,\theta = \frac{{\left| {{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}} \right|}}{{\sqrt {a_1^2 + b_1^2 + c_1^2} .\sqrt {a_2^2 + b_2^2 + c_2^2} }} \cr & \Rightarrow \cos \,\theta = 0 \cr & \Rightarrow \theta = \frac{\pi }{2} \cr} $$

$$\eqalign{ & AC = \overrightarrow {AB} .\widehat {AC} = \left( {\hat i + \hat j} \right).\frac{{\left( {\hat i + \hat j + \hat k} \right)}}{{\sqrt 3 }} = \frac{2}{{\sqrt 3 }} \cr & {\text{Now, }}A'B' = BC = \sqrt {A{B^2} - A{C^2}} \, = \sqrt {2 - \frac{4}{3}} = \sqrt {\frac{2}{3}} \, \cr} $$
$$\therefore $$ Length of projection $$ = \sqrt {\frac{2}{3}} \,$$

The equation of the given sphere is
$${x^2} + {y^2} + {z^2} - 10z = 0.$$
$$\therefore $$  Its centre is $$\left( {0,\,0,\,5} \right).$$
Coordinates of one end point of a diameter of the sphere is given as $$\left( { - 3,\, - 4,\,5} \right).$$
Let Coordinates of another end point of this diameter $$\left( {{x_1},\,{y_1},\,{z_1}} \right).$$
$$\eqalign{ & \therefore \,\frac{{ - 3 + {x_1}}}{2} = 0 \Rightarrow {x_1} = 3, \cr & \frac{{ - 4 + {y_1}}}{2} = 0 \Rightarrow {y_1} = 4 \cr & {\text{and }}\frac{{5 + {z_1}}}{2} = 5 \Rightarrow {z_1} = 5 \cr} $$
$$\therefore $$  Required coordinates are $$\left( {3,\,4,\,5} \right).$$

Given d’c’s of two lines are $$l,\,m,\,n$$  connected by the relations $$l + m + n = 0$$   and $$lm = 0$$
$$\eqalign{ & {\text{Now, }}l + m + n = 0 \cr & \Rightarrow l = - m - n \cr & \Rightarrow l = - \left( {m + n} \right) \cr & {\text{and }}lm = 0 \cr & \Rightarrow - \left( {m + n} \right)m = 0 \cr & \Rightarrow - mm - mn = 0 \cr & \Rightarrow mm = - mn\,; \cr & {\text{Therefore }}m{\text{ and }}m + n = 0 \cr & {\text{Then }}\frac{{{l_1}}}{{ - 1}} = \frac{{{m_1}}}{0} = \frac{{{n_1}}}{1}\,{\text{and if }}l + m + n = 0 \cr & {\text{then }}\frac{{{l_2}}}{0} = \frac{{{m_2}}}{{ - 1}} = \frac{{{n_2}}}{1}\, \cr & \left( {{l_1},\,{m_1},\,{n_1}} \right) = \left( { - 1,\,0,\,1} \right)\,{\text{and }}\left( {{l_2},\,{m_2},\,{n_2}} \right) = \left( {0,\, - 1,\,1} \right) \cr} $$
We know that angle between them
$$\eqalign{ & \cos \,\theta = \frac{{0 + 0 + 1}}{{\sqrt {1 + 0 + 1} \sqrt {0 + 1 + 1} }} \cr & \Rightarrow \cos \,\theta = \frac{1}{{\sqrt 2 \sqrt 2 }} \cr & \Rightarrow \cos \,\theta = \frac{1}{2} \cr & \Rightarrow \cos \,\theta = \cos \,{60^ \circ } \cr & \Rightarrow \theta = {60^ \circ } \cr & \Rightarrow \theta = \frac{\pi }{3} \cr} $$

As perpendicular distance of $$x + 2y - 2z = \alpha $$    from the point $$\left( {1,\, - 2,\,1} \right)$$   is 5

$$\eqalign{ & \therefore \left| {\frac{{1 - 4 - 2 - \alpha }}{3}} \right| = 5 \cr & \Rightarrow \frac{{ - 5 - \alpha }}{3} = 5{\text{ or }} - 5 \cr & \Rightarrow \alpha = - 20{\text{ or }}10 \cr & {\text{As }}\alpha > 0\,\,\, \Rightarrow \alpha = 10 \cr} $$
$$\therefore $$ Plane becomes $$x+2y-2z-10=0$$
Equation of PN is $$\frac{{x - 1}}{1} = \frac{{y + 2}}{2} = \frac{{z - 1}}{{ - 2}} = \lambda $$
For some value of $$\lambda ,\,\,\,\,{\text{N}}\left( {\lambda + 1,\,2\lambda - 2,\, - 2\lambda + 1} \right)$$
It lies on $$x+2y-2z-10=0$$
$$\eqalign{ & \therefore \lambda + 1 + 4\lambda - 4 + 4\lambda - 2 = 10 \cr & \Rightarrow 9\lambda = 15 \cr & \Rightarrow \lambda = \frac{5}{3} \cr & \therefore \,\,{\text{N}}\left( {\frac{8}{3},\,\frac{4}{3},\,\frac{{ - 7}}{3}} \right) \cr} $$