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

Let the area of the triangle formed by the lines $x+2=y-1=z$,$\frac{x-3}{5}=\frac{y}{-1}=\frac{z-1}{1}$ and $\frac{x}{-3}=\frac{y-3}{3}=\frac{z-2}{1}$ be $A$. Then $A^2$ is equal to . . . . . . .

If the lines $\frac{3-x}{2}=\frac{5y-2}{3\lambda+1}=5-z$ and $\frac{x+2}{-1}=\frac{1-3y}{7}=\frac{4-z}{2\mu}$ are at right angles,then $7\lambda-10\mu=$

If $\frac{x - 1}{l} = \frac{y - 2}{m} = \frac{z + 1}{n}$ is the equation of the line passing through the points $(1, 2, -1)$ and $(-1, 0, 1)$,then the direction ratios $(l, m, n)$ are:

If the length of the perpendicular drawn from the point $P(a, 4, 2)$,$a > 0$ on the line $\frac{x+1}{2} = \frac{y-3}{3} = \frac{z-1}{-1}$ is $2\sqrt{6}$ units and $Q(\alpha_{1}, \alpha_{2}, \alpha_{3})$ is the image of the point $P$ in this line,then $a + \sum_{i=1}^{3} \alpha_{i}$ is equal to.

Find the point of intersection of the lines $\frac{x - 4}{5} = \frac{y - 1}{2} = \frac{z}{1}$ and $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$.