If the line joining the points $(k, 2, 3)$ and $(1, 1, 2)$ is parallel to the line joining the points $(5, 4, -1)$ and $(3, 2, -3)$,then the value of $k$ is equal to

Let $l_1$ be the line passing through the point $A = 3\hat{i} + 4\hat{j} - 2\hat{k}$ and parallel to the vector $\vec{b_1} = -\hat{i} + 2\hat{j} + \hat{k}$. Let $l_2$ be another line passing through the point $B = \hat{i} - 7\hat{j} - 2\hat{k}$ and parallel to the vector $\vec{b_2} = \hat{i} + 3\hat{j} + 2\hat{k}$. Then the shortest distance between the lines $l_1$ and $l_2$ is:

Find the coordinates of the foot of the perpendicular from the point $(2, 4, 1)$ to the line $\vec{r} = (-5, -3, 6) + k(1, 4, -9)$,where $k \in R$.

The shortest distance between the skew lines $\vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+t(\hat{i}+3 \hat{j}+2 \hat{k})$ and $\vec{r}=(4 \hat{i}+5 \hat{j}+6 \hat{k})+s(2 \hat{i}+3 \hat{j}+\hat{k})$ is

If the shortest distance between the lines $\vec{r}=(-\hat{i}+3\hat{k})+\lambda(\hat{i}-a\hat{j})$ and $\vec{r}=(-\hat{j}+2\hat{k})+\mu(\hat{i}-\hat{j}+\hat{k})$ is $\sqrt{\frac{2}{3}}$,then the integral value of $a$ is equal to

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: