The length of the perpendicular from the point $A(1, -2, -3)$ on the line $\frac{x-1}{2} = \frac{y+3}{-1} = \frac{z+1}{-2}$ is (in $\text{ units}$)

The two lines $L_1: \vec{r}=(\hat{i}+5 \hat{j}+5 \hat{k})+t(4 \hat{i}-4 \hat{j}+5 \hat{k})$ and $L_2: \vec{r}=(2 \hat{i}+4 \hat{j}+5 \hat{k})+s(8 \hat{i}-3 \hat{j}+\hat{k})$ are such that

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 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

Let $(\alpha, \beta, \gamma)$ be the mirror image of the point $(2, 3, 5)$ in the line $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$. Then $2\alpha + 3\beta + 4\gamma$ is equal to

If the line passing through the points $(a, 1, 6)$ and $(3, 4, b)$ crosses the $yz$-plane at the point $\left(0, \frac{17}{2}, \frac{-13}{2}\right)$,then: