Find the angle between the lines $\frac{x - 2}{3} = \frac{y + 1}{-2}; z = 2$ and $\frac{x - 1}{1} = \frac{2y + 3}{3} = \frac{z + 5}{2}$.

$A$ line with direction cosines proportional to $2, 1, 2$ meets the line $L_1$ passing through $(0, -1, 0)$ with direction ratios $1, 1, 1$ at $A(x, y, z)$ and another line $L_2$ at $B(1, 1, 1)$. Then $x+y+z=$

The length of the perpendicular drawn from the point $2 \hat{i} - \hat{j} + 5 \hat{k}$ to the line $\vec{r} = (11 \hat{i} - 2 \hat{j} - 8 \hat{k}) + \lambda(10 \hat{i} - 4 \hat{j} - 11 \hat{k})$ 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

The straight line $\frac{x - 3}{3} = \frac{y - 2}{1} = \frac{z - 1}{0}$ is

If the image of the point $P(1, 2, a)$ in the line $\frac{x-6}{3}=\frac{y-7}{2}=\frac{7-z}{2}$ is $Q(5, b, c)$, then $a^{2}+b^{2}+c^{2}$ is equal to