Let $Q$ be the foot of the perpendicular from the point $P(7, -2, 13)$ on the plane containing the lines $\frac{x+1}{6} = \frac{y-1}{7} = \frac{z-3}{8}$ and $\frac{x-1}{3} = \frac{y-2}{5} = \frac{z-3}{7}$. Then $(PQ)^{2}$ is equal to ..... .

The distance of the point $P(3, 8, 2)$ from the line $\frac{x-1}{2} = \frac{y-3}{4} = \frac{z-2}{3}$ measured parallel to the plane $3x + 2y - 2z + 15 = 0$ is (in $\text{ units}$)

Statement-$1$: The point $A(1, 0, 7)$ is the reflection of the point $B(1, 6, 3)$ in the line $\frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3}$.
Statement-$2$: The line $\frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3}$ is the perpendicular bisector of the line segment joining $A(1, 0, 7)$ and $B(1, 6, 3)$.

The direction cosines of the line formed by the intersection of the planes $x - y + 2z = 5$ and $3x + y + z = 6$ are:

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 equation of the plane through the intersection of the planes $x+2y+3z-4=0$ and $4x+3y+2z+1=0$ and passing through the origin is