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.) $$-2$$

Two lines $${L_1}:x = 5,{y \over {3 - \alpha }} = {z \over { - 2}}$$ and $${L_2}:x = \alpha ,{y \over { - 1}} = {z \over {2 - \alpha }}$$ are coplanar. Then $$\alpha $$ can take value(s)

A box $${B_1}$$ contains $$1$$ white ball, $$3$$ red balls and $$2$$ black balls. Another box $${B_2}$$ contains $$2$$ white balls, $$3$$ red balls and $$4$$ black balls. A third box $${B_3}$$ contains $$3$$ white balls, $$4$$ red balls and $$5$$ black balls.

If $$1$$ ball is drawn from each of the boxex $${B_1},$$ $${B_2}$$ and $${B_3},$$ the probability that all $$3$$ drawn balls are of the same colour is

A box $${B_1}$$ contains $$1$$ white ball, $$3$$ red balls and $$2$$ black balls. Another box $${B_2}$$ contains $$2$$ white balls, $$3$$ red balls and $$4$$ black balls. A third box $${B_3}$$ contains $$3$$ white balls, $$4$$ red balls and $$5$$ black balls.

If $$2$$ balls are drawn (without replacement) from a randomly selected box and one of the balls is white and the other ball is red, the probability that these $$2$$ balls are drawn from box $${B_2}$$ is