Let $P$ and $Q$ be the points on the line $\frac{x+3}{8}=\frac{y-4}{2}=\frac{z+1}{2}$ which are at a distance of $6$ units from the point $R(1,2,3)$. If the centroid of the triangle $PQR$ is $(\alpha, \beta, \gamma)$,then $\alpha^2+\beta^2+\gamma^2$ is:

Find the coordinates of the foot of the perpendicular drawn from the point $A(1, 0, 3)$ to the line joining the points $B(4, 7, 1)$ and $C(3, 5, 3)$.

The shortest distance between the lines $r = (3i - 2j - 2k) + t(i)$ and $r = (i - j + 2k) + s(j)$ ($t$ and $s$ being parameters) is

The equation of a line passing through the point $(-1, 2, 3)$ and perpendicular to the lines $\frac{x}{2} = \frac{y-1}{-3} = \frac{z+2}{-2}$ and $\frac{x+3}{-1} = \frac{y+3}{2} = \frac{z-1}{3}$ is

Let the image of the point $P(1, 2, 3)$ in the line $L : \frac{x-6}{3} = \frac{y-1}{2} = \frac{z-2}{3}$ be $Q$. Let $R(\alpha, \beta, \gamma)$ be a point that divides the line segment $PQ$ internally in the ratio $1:3$. Then the value of $22(\alpha+\beta+\gamma)$ is equal to

The lines whose direction cosines satisfy the equations $al + bm + cn = 0$ and $fmn + gnl + hlm = 0$ are perpendicular if: