The direction ratios of the line given by the intersection of the planes $x - y + z - 5 = 0$ and $x - 3y - 6 = 0$ are:

The distance of the point $(1, -2, 4)$ from the plane passing through the point $(1, 2, 2)$ and perpendicular to the planes $x - y + 2z = 3$ and $2x - 2y + z + 12 = 0$ is:

Let the line $\frac{x - 2}{3} = \frac{y - 1}{-5} = \frac{z + 2}{2}$ lie in the plane $x + 3y - \alpha z + \beta = 0$. Then $(\alpha, \beta)$ equals.

Let $P$ be the plane $\sqrt{3} x+2 y+3 z=16$ and let $S=\left\{\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}: \alpha^2+\beta^2+\gamma^2=1 \text{ and the distance of } (\alpha, \beta, \gamma) \text{ from the plane } P \text{ is } \frac{7}{2}\right\}$. Let $\overrightarrow{u}, \overrightarrow{v}$ and $\overrightarrow{w}$ be three distinct vectors in $S$ such that $|\overrightarrow{u}-\overrightarrow{v}|=|\overrightarrow{v}-\overrightarrow{w}|=|\overrightarrow{w}-\overrightarrow{u}|$. Let $V$ be the volume of the parallelepiped determined by vectors $\overrightarrow{u}, \overrightarrow{v}$ and $\overrightarrow{w}$. Then the value of $\frac{80}{\sqrt{3}} V$ is

If the shortest distance between the lines $\frac{x - 1}{\alpha} = \frac{y + 1}{-1} = \frac{z}{1}, (\alpha \ne -1)$ and $x + y + z + 1 = 0 = 2x - y + z + 3$ is $\frac{1}{\sqrt{3}}$, then a value of $\alpha$ is

Let a plane $P$ pass through the point $(3, 7, -7)$ and contain the line $\frac{x-2}{-3} = \frac{y-3}{2} = \frac{z+2}{1}$. If the distance of the plane $P$ from the origin is $d$,then $d^{2}$ is equal to $.....$