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 :

Let a straight line $L$ pass through the point $P(2, -1, 3)$ and be perpendicular to the lines $\frac{x-1}{2} = \frac{y+1}{1} = \frac{z-3}{-2}$ and $\frac{x-3}{1} = \frac{y-2}{3} = \frac{z+2}{4}$. If the line $L$ intersects the $yz$-plane at the point $Q$,then the distance between the points $P$ and $Q$ is:

The equation of the line of shortest distance between the lines $\frac{x - 6}{3} = \frac{y - 7}{-1} = \frac{z - 4}{1}$ and $\frac{x}{-3} = \frac{y + 9}{2} = \frac{z - 2}{4}$ is:

The shortest distance between lines $\bar{r}=(\hat{i}+2 \hat{j}-\hat{k})+\lambda(2 \hat{i}+\hat{j}-3 \hat{k})$ and $\bar{r}=(2 \hat{i}-\hat{j}+2 \hat{k})+\mu(\hat{i}-\hat{j}+\hat{k})$ is

The square of the distance of the point $P(5,6,7)$ from the line $\frac{x-2}{2} = \frac{y-5}{3} = \frac{z-2}{4}$ is equal to:

Which of the following lines in ${R^3}$ is coincident with the line $\vec{r} = (1, 1, 1) + k(2, 3, 4), k \in R$?