Consider the lines

$${L_1}:{{x - 1} \over 2} = {y \over { - 1}} = {{z + 3} \over 1},{L_2} : {{x - 4} \over 1} = {{y + 3} \over 1} = {{z + 3} \over 2}$$

and the planes $${P_1}:7x + y + 2z = 3,{P_2} = 3x + 5y - 6z = 4.$$ Let $$ax+by+cz=d$$ be the equation of the plane passing through the point of intersection of lines $${L_1}$$ and $${L_2},$$ and perpendicular to planes $${P_1}$$ and $${P_2}.$$

Match List $$I$$ with List $$II$$ and select the correct answer using the code given below the lists:
List $$I$$
(P.) $$a=$$
(Q.) $$b=$$
(R.) $$c=$$
(S.) $$d=$$

List $$II$$
(1.) $$13$$
(2.) $$-3$$
(3.) $$1$$
(4.) $$-2$$

Perpendiculars are drawn from points on the line $\frac{x+2}{2}=\frac{y+1}{-1}=\frac{z}{3}$ to the plane $x+y+$ $z=3$. The foot of perpendiculars lie on the line

The equation of a plane passing through the line of intersection of the planes $$x+2y+3z=2$$ and $$x-y+z=3$$ and at a distance $${2 \over {\sqrt 3 }}$$ from the point $$(3, 1, -1)$$ is

The point $$P$$ is the intersection of the straight line joining the points $$Q(2, 3, 5)$$ and $$R(1, -1, 4)$$ with the plane $$5x-4y-z=1.$$ If $$S$$ is the foot of the perpendicular drawn from the point $$T(2, 1, 4)$$ to $$QR,$$ then the length of the line segment $$PS$$ is