Let $d$ be the distance between the foot of perpendiculars of the points $P(1, 2, -1)$ and $Q(2, -1, 3)$ on the plane $-x + y + z = 1$. Then $d^{2}$ 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 $\text{ units}$)

If $L$ is a line common to the planes $3x + 4y + 7z = 1$ and $x - y + z = 5$,then the direction ratios of the line $L$ are:

If the straight lines $x = 1 + s, y = -3 - \lambda s, z = 1 + \lambda s$ and $x = t/2, y = 1 + t, z = 2 - t$,with parameters $s$ and $t$ respectively,are coplanar,then $\lambda$ equals

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

Find the coordinates of the point where the line passing through $(3, -4, -5)$ and $(2, -3, 1)$ intersects the plane $2x + y + z = 7$.