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 $.....$

The line of intersection of the planes $r \cdot (i - 3j + k) = 1$ and $r \cdot (2i + 5j - 3k) = 2$ is parallel to the vector

$A$ perpendicular is drawn from a point $P$ on the line $\frac{x - 1}{2} = \frac{y + 1}{-1} = \frac{z}{1}$ to the plane $x + y + z = 3$ such that the foot of the perpendicular $Q$ also lies on the plane $x - y + z = 3$. Then the coordinates of $Q$ are

$A$ line drawn from the point $A(1, 3, 2)$ parallel to the line $\frac{x}{2} = \frac{y}{4} = \frac{z}{1}$ intersects the plane $3x + y + 2z = 5$ at point $B$. Then the coordinates of point $B$ are:

If the angle between the line $x = \frac{y-1}{2} = \frac{z-3}{\lambda}$ and the plane $x + 2y + 3z = 4$ is $\cos^{-1} \sqrt{\frac{5}{14}}$,then the value of $\lambda$ is

The lines $\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-k}$ and $\frac{x-1}{k}=\frac{y-4}{2}=\frac{z-5}{1}$ are coplanar if