Consider the lines,

$${L_1}:{{x + 1} \over 3} = {{y + 2} \over 1} = {{z + 1} \over 2}$$

$${L_2}:{{x - 2} \over 1} = {{y - 2} \over 2} = {{z - 3} \over 3}$$

The distance of the point $$(1, 1, 1)$$ from the plane passing through the point $$(-1, -2, -1)$$ and whose normal is perpendicular to both the lines $${L_1}$$ and $${L_2}$$ is :

Consider three planes $$${P_1}:x - y + z = 1$$$ $$${P_2}:x + y - z = 1$$$ $$${P_3}:x - 3y + 3z = 2$$$

Let $${L_1},$$ $${L_2},$$ $${L_3}$$ be the lines of intersection of the planes $${P_2}$$ and $${P_3},$$ $${P_3}$$ and $${P_1},$$ $${P_1}$$ and $${P_2},$$ respectively.

STATEMENT - 1Z: At least two of the lines $${L_1},$$ $${L_2}$$ and $${L_3}$$ are non-parallel and

STATEMENT - 2: The three planes doe not have a common point.

Consider the planes $$3 x-6 y-2 z=15$$ and $$2 x+y-2 z=5$$.

STATEMENT - 1 : The parametric equations of the line of intersection of the given planes are $$x=3+14 t, y=1+2 t, z=15 t$$

STATEMENT - 2 : The vectors $$14 \hat{i}+2 \hat{j}+15 \hat{k}$$ is parallel to the line of intersection of the given planes.

Match the following:

(i)	$$\sum\limits_{i = 1}^\infty {{{\tan }^{ - 1}}\left( {{1 \over {2{i^2}}}} \right) = t} $$ then $$\tan t=$$	(A)	0
(ii)	Sides $$a,b,c$$ of a triangle ABC are in AP and $$\cos {\theta _1} = {a \over {b + c}},\cos {\theta _2} = {b \over {a + c}},\cos {\theta _3} = {c \over {a + b}}$$, then $${\tan ^2}\left( {{{{\theta _1}} \over 2}} \right) + {\tan ^2}\left( {{{{\theta _3}} \over 2}} \right) = $$	(B)	1
(iii)	A line is perpendicular to $$x + 2y + 2z = 0$$ and passes through (0, 1, 0). The perpendicular distance of this line from the origin is	(C)	$${{\sqrt 5 } \over 3}$$
		(D)	2/3