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

The coordinates of the foot of the perpendicular from the point $(0,2,3)$ on the line $\frac{x+3}{5}=\frac{y+1}{2}=\frac{z+4}{3}$ are

The equation of the line joining the points $(-3, 4, 11)$ and $(1, -2, 7)$ is

Line $L_1$ passes through the points $\hat{i}+\hat{j}$ and $\hat{k}-\hat{i}$. Line $L_2$ passes through the point $\hat{j}+2\hat{k}$ and is parallel to the vector $\hat{i}+\hat{j}+\hat{k}$. If $x\hat{i}+y\hat{j}+z\hat{k}$ is the point of intersection of the lines $L_1$ and $L_2$,then $(y-x)=$

If the direction ratios of two lines are given by $3lm - 4ln + mn = 0$ and $l + 2m + 3n = 0$,what is the angle between them?

The length of the perpendicular from the point $P(2, -1, 4)$ to the straight line $\frac{x + 3}{10} = \frac{y - 2}{-7} = \frac{z}{1}$ is