The shortest distance between the lines $\frac{x-5}{1}=\frac{y-2}{2}=\frac{z-4}{-3}$ and $\frac{x+3}{1}=\frac{y+5}{4}=\frac{z-1}{-5}$ is (in $\sqrt{3}$)

Let $ABC$ be a triangle with $A(\alpha, 5, \beta)$,$B(-2, 1, 6)$ and $C(1, 0, -3)$ as its vertices. If the median through $B$ is equally inclined to the coordinate axes,then $\alpha + \beta =$

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,passing through $A(1, 2, 3)$ and perpendicular to the vectors $2 \hat{i} + \hat{j} - \hat{k}$ and $\hat{i} + 3 \hat{j} + 2 \hat{k}$,is

The lines $x = ay + b, z = cy + d$ and $x = a'y + b', z = c'y + d'$ are perpendicular to each other,if

Find the shortest distance between the lines given by $\vec{r}=(8+3 \lambda) \hat{i}+(-9-16 \lambda) \hat{j}+(10+7 \lambda) \hat{k}$ and $\vec{r}=15 \hat{i}+29 \hat{j}+5 \hat{k}+\mu(3 \hat{i}+8 \hat{j}-5 \hat{k})$. (in $\text{ units}$)