The line $l_1$ passes through the point $(2, 6, 2)$ and is perpendicular to the plane $2x + y - 2z = 10$. Then the shortest distance between the line $l_1$ and the line $\frac{x + 1}{2} = \frac{y + 4}{-3} = \frac{z}{2}$ is :

The angle between the lines $\bar{r}=(\hat{i}+2\hat{j}+3\hat{k})+\lambda(\hat{i}+\hat{j}+2\hat{k})$ and $\bar{r}=(3\hat{i}+\hat{k})+\lambda^{\prime}(2\hat{i}+\hat{j}-\hat{k})$,where $\lambda, \lambda^{\prime} \in R$,is

If the shortest distance between the lines $r=(3 \hat{i}+4 \hat{j}-2 \hat{k})+t(-\hat{i}+2 \hat{j}+\hat{k})$ and $r=(\hat{i}-7 \hat{j}-2 \hat{k})+s(\hat{i}+3 \hat{j}+2 \hat{k})$ is equivalent to the projection of $P=-2 \hat{i}+11 \hat{j}$ on $Q$, then a possible vector $Q$ is

Let $l_{1}$ be the line in $xy$-plane with $x$ and $y$ intercepts $\frac{1}{8}$ and $\frac{1}{4 \sqrt{2}}$ respectively,and $l_{2}$ be the line in $zx$-plane with $x$ and $z$ intercepts $-\frac{1}{8}$ and $-\frac{1}{6 \sqrt{3}}$ respectively. If $d$ is the shortest distance between the line $l_{1}$ and $l_{2}$,then $d^{-2}$ is equal to

The perpendicular distance from the point $P(3, 5, 2)$ to the line $L$ passing through the point $2\hat{i} + \hat{j}$ and parallel to the vector $\hat{i} + 5\hat{j} + 2\hat{k}$ is

$P, Q, R$ and $S$ are four points with the position vectors $3i-4j+5k, 0i+0j+4k, -4i+5j+1k$ and $-3i+4j+3k$,respectively. Then,the line $PQ$ meets the line $RS$ at the point