The ratio in which the line segment joining the points $(2, -4, 3)$ and $(-4, 5, -6)$ is divided by the plane $3x + 2y + z - 4 = 0$ is:

The point of intersection of the line $x+1=\frac{y+3}{3}=\frac{-z+2}{2}$ with the plane $3x+4y+5z=10$ is

The equation of the line passing through the point $(2, 3, 1)$ and parallel to the line of intersection of the planes $x - 2y - z + 5 = 0$ and $x + y + 3z = 6$ is:

If the equation of a line is $\frac{x + 3}{2} = \frac{y - 4}{3} = \frac{z + 5}{2}$ and the equation of a plane is $4x - 2y - z = 1$,then which of the following is true?

For a positive real number $p$,if the perpendicular distance from a point $-\hat{i} + p\hat{j} - 3\hat{k}$ to the plane $\vec{r} \cdot (2\hat{i} - 3\hat{j} + 6\hat{k}) = 7$ is $6$ units,then $p=$

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	$(4) \frac{1}{\sqrt{6}} \hat{i}-\frac{2}{\sqrt{6}} \hat{j}+\frac{1}{\sqrt{6}} \hat{k}$
	$(5) \sqrt{\frac{2}{3}}$