Let $(\alpha, \beta, \gamma)$ be the foot of the perpendicular from the point $(1, 2, 3)$ on the line $\frac{x+3}{5} = \frac{y-1}{2} = \frac{z+4}{3}$. Then $19(\alpha + \beta + \gamma)$ is equal to:

The sum of all values of $ \alpha $, for which the shortest distance between the lines $ \frac{x+1}{\alpha}=\frac{y-2}{-1}=\frac{z-4}{-\alpha} $ and $ \frac{x}{\alpha}=\frac{y-1}{2}=\frac{z-1}{2\alpha} $ is $ \sqrt{2} $, is

$A$ 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 angle between two lines $\frac{x+1}{2}=\frac{y+3}{2}=\frac{z-4}{-1}$ and $\frac{x-4}{1}=\frac{y+4}{2}=\frac{z+1}{2}$ is

If the direction ratios of two lines are given by $3lm - 4ln + mn = 0$ and $l + 2m + 3n = 0$,then the angle between the lines is

The lines $\bar{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(3 \hat{i} - \hat{j})$ and $\bar{r} = (4 \hat{i} - \hat{k}) + \mu(2 \hat{i} + 3 \hat{k})$ are