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 :

Consider the lines $L_1$ and $L_2$ given by
$L_1: \frac{x-1}{2} = \frac{y-3}{1} = \frac{z-2}{2}$
$L_2: \frac{x-2}{1} = \frac{y-2}{2} = \frac{z-3}{3}$
$A$ line $L_3$ having direction ratios $1, -1, -2$ intersects $L_1$ and $L_2$ at the points $P$ and $Q$ respectively. Then the length of line segment $PQ$ is

If points $P(4, 5, x)$,$Q(3, y, 4)$,and $R(5, 8, 0)$ are collinear,then the value of $x+y$ is

The distance of the point $(2, 3, 4)$ from the line $1 - x = \frac{y}{2} = \frac{1}{3}(1 + z)$ is

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

Find the vector and the cartesian equations of the line that passes through the points $(3,-2,-5)$ and $(3,-2,6).$