The equation of the plane passing through the point (1, 1, 1) and perpendicular to the planes 2x + y $$-$$ 2z = 5 and 3x $$-$$ 6y $$-$$ 2z = 7 is

Let $$P$$ be the image of the point $$(3,1,7)$$ with respect to the plane $$x-y+z=3.$$ Then the equation of the plane passing through $$P$$ and containing the straight line $${x \over 1} = {y \over 2} = {z \over 1}$$ is

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