$A$ plane $\pi$ passing through the points $2 \hat{i}-3 \hat{j}$ and $3 \hat{i}+4 \hat{k}$ is parallel to the vector $2 \hat{i}+3 \hat{j}-4 \hat{k}$. If a line joining the points $\hat{i}+2 \hat{j}$ and $\hat{j}-2 \hat{k}$ intersects the plane $\pi$ at the point $a \hat{i}+b \hat{j}+c \hat{k}$, then $a+b+2c=$

Let $Q$ and $R$ be the feet of perpendiculars from the point $P(a, a, a)$ on the lines $x=y, z=1$ and $x=-y, z=-1$ respectively. If $\angle QPR$ is a right angle,then $12a^2$ is equal to

Show that the planes $\vec{r} \cdot (\hat{i} + 2\hat{j} + 2\hat{k}) = 19$ and $\vec{r} \cdot (4\hat{i} - 3\hat{j} + 12\hat{k}) + 3 = 0$ are perpendicular. Find the equation of the plane containing these two lines (Note: The question asks for the plane containing the intersection of these two planes).

Let the equation of the plane passing through the line of intersection of the planes $x+2y+az=2$ and $x-y+z=3$ be $5x-11y+bz=6a-1$. For $c \in \mathbb{Z}$,if the distance of this plane from the point $(a, -c, c)$ is $\frac{2}{\sqrt{a}}$,then $\frac{a+b}{c}$ is equal to

The distance of the origin from the plane $r \cdot(3 \hat{i}+4 \hat{j}-12 \hat{k})=7$ measured parallel to the line $r=(\hat{i}+2 \hat{j}+3 \hat{k})+t(6 \hat{i}+2 \hat{j}+3 \hat{k})$ is

Let $P_1$ and $P_2$ be two planes given by $P_1: 10x + 15y + 12z - 60 = 0$ and $P_2: -2x + 5y + 4z - 20 = 0$. Which of the following straight lines can be an edge of some tetrahedron whose two faces lie on $P_1$ and $P_2$?
$(A) \frac{x-1}{0} = \frac{y-1}{0} = \frac{z-1}{5}$
$(B) \frac{x-6}{-5} = \frac{y}{2} = \frac{z}{3}$
$(C) \frac{x}{-2} = \frac{y-4}{5} = \frac{z}{4}$
$(D) \frac{x}{1} = \frac{y-4}{-2} = \frac{z}{3}$