The line passing through the points $(a, 1, 6)$ and $(3, 4, b)$ crosses the $yz$-plane at $\left(0, \frac{17}{2}, \frac{-13}{2}\right)$. Then the value of $(3a + 4b)$ is:

The shortest distance (in units) between the lines $\frac{x+1}{3}=\frac{y+2}{1}=\frac{z+1}{2}$ and $\vec{r}=(2\hat{i}-2\hat{j}+3\hat{k})+\lambda(\hat{i}+2\hat{j})$ is

The shortest distance between the lines $\frac{x-2}{3}=\frac{y+1}{2}=\frac{z-6}{2}$ and $\frac{x-6}{3}=\frac{1-y}{2}=\frac{z+8}{0}$ is equal to $............$

Let the point $A$ be the foot of the perpendicular drawn from the point $P(a, b, 0)$ to the line $\frac{x-1}{2} = \frac{y-2}{1} = \frac{z-\alpha}{3}$. If the midpoint of the line segment $PA$ is $(0, \frac{3}{4}, -\frac{1}{4})$, then the value of $a^2+b^2+\alpha^2$ is equal to:

Let $A(\alpha, 4, 7)$ and $B(3, \beta, 8)$ be two points in space. If the $YZ$ plane and $ZX$ plane respectively divide the line segment joining the points $A$ and $B$ in the ratio $2:3$ and $4:5$, then the point $C$ which divides $\overline{AB}$ in the ratio $\alpha: \beta$ externally is

If the shortest distance between the line joining the points $(1, 2, 3)$ and $(2, 3, 4)$,and the line $\frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-2}{0}$ is $\alpha$,then $28 \alpha^2$ is equal to $........$.