Find the equation of the plane perpendicular to the plane containing the line $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$ and the lines $\frac{x}{3} = \frac{y}{4} = \frac{z}{2}$ and $\frac{x}{4} = \frac{y}{2} = \frac{z}{3}$.

Let $\gamma \in R$ be such that the lines $L_1: \frac{x+11}{1}=\frac{y+21}{2}=\frac{z+29}{3}$ and $L_2: \frac{x+16}{3}=\frac{y+11}{2}=\frac{z+4}{\gamma}$ intersect. Let $R_1$ be the point of intersection of $L_1$ and $L_2$. Let $O=(0,0,0)$,and $\hat{n}$ denote a unit normal vector to the plane containing both the lines $L_1$ and $L_2$. Match each entry in $List-I$ to the correct entry in $List-II$.

$List-I$	$List-II$
$(P) \gamma$ equals	$(1) -\hat{i}-\hat{j}+\hat{k}$
$(Q)$ $A$ possible choice for $\hat{n}$ is	$(2) \sqrt{\frac{3}{2}}$
$(R) \vec{OR_1}$ equals	$(3) 1$
$(S)$ $A$ possible value of $\vec{OR_1} \cdot \hat{n}$ is	$(4) \frac{1}{\sqrt{6}} \hat{i}-\frac{2}{\sqrt{6}} \hat{j}+\frac{1}{\sqrt{6}} \hat{k}$
	$(5) \sqrt{\frac{2}{3}}$

If the line $\frac{x - x_1}{l} = \frac{y - y_1}{m} = \frac{z - z_1}{n}$ is parallel to the plane $ax + by + cz + d = 0$,then:

The plane $XOZ$ divides the join of $(1, -1, 5)$ and $(2, 3, 4)$ in the ratio $\lambda :1$. Then,the value of $\lambda$ is:

The line of intersection of the planes $r \cdot (i - 3j + k) = 1$ and $r \cdot (2i + 5j - 3k) = 2$ is parallel to which vector?

If the three planes $x = 5, 2x - 5ay + 3z - 2 = 0$ and $3bx + y - 3z = 0$ contain a common line,then $(a, b)$ is equal to