For what value of $c$ does the line $\frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 1}{-1}$ intersect the curve $xy = c^2, z = 0$?

The distance of the point $(2, 3, -5)$ from the plane $x + 2y - 2z = 9$ is

Let the line $\ell: x = \frac{1-y}{-2} = \frac{z-3}{\lambda}, \lambda \in R$ meet the plane $P: x + 2y + 3z = 4$ at the point $(\alpha, \beta, \gamma)$. If the angle between the line $\ell$ and the plane $P$ is $\cos^{-1}\left(\sqrt{\frac{5}{14}}\right)$,then $\alpha + 2\beta + 6\gamma$ is equal to

The equation of the plane passing through the intersection of the planes $x+y+z=1$ and $2x+3y-z+4=0$ and parallel to the $X$-axis is

If the product of distances of the point $(1, 1, 1)$ from the origin and the plane $x - y + z + k = 0$ is $5$,then $k =$

$A$ line with direction cosines passes through the point $P(2, -1, 2)$ and makes equal angles with the coordinate axes. If the line meets the plane $2x + y + z = 9$ at point $Q$,then the length of $PQ$ is . . . . . . .