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

The line $\frac{x-1}{2}=\frac{y+2}{-1}=\frac{z}{1}$ intersects the $XY$ plane and the $YZ$ plane at points $A$ and $B$ respectively. The equation of the line passing through the points $A$ and $B$ is

$A$ plane $ax+by+cz+1=0$ is perpendicular to the two planes $2x-2y+z=0$ and $x-y+2z=4$ and passes through the point $(1, -2, 1)$. Then $a+b-c=$

Let the lines $L_1: \vec{r}=\hat{i}+2\hat{j}+3\hat{k}+\lambda(2\hat{i}+3\hat{j}+4\hat{k})$,$\lambda \in R$ and $L_2: \vec{r}=(4\hat{i}+\hat{j})+\mu(5\hat{i}+2\hat{j}+\hat{k})$,$\mu \in R$,intersect at the point $R$. Let $P$ and $Q$ be the points lying on lines $L_1$ and $L_2$,respectively,such that $|\overrightarrow{PR}|=\sqrt{29}$ and $|\overrightarrow{PQ}|=\sqrt{\frac{47}{3}}$. If the point $P$ lies in the first octant,then $27(QR)^2$ is equal to

The image of the point with position vector $(\hat{i}+3 \hat{j}+4 \hat{k})$ in the plane $r \cdot(2 \hat{i}-\hat{j}+\hat{k})+3=0$ is

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 $p=$