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 $7\lambda-10\mu=$

Let $(\alpha, \beta, \gamma)$ be the coordinates of the foot of the perpendicular drawn from the point $(5, 4, 2)$ on the line $\vec{r} = (-\hat{i} + 3\hat{j} + \hat{k}) + \lambda(2\hat{i} + 3\hat{j} - \hat{k})$. Then the length of the projection of the vector $\alpha\hat{i} + \beta\hat{j} + \gamma\hat{k}$ on the vector $6\hat{i} + 2\hat{j} + 3\hat{k}$ is:

If the lines $\frac{x-1}{-3}=\frac{y-2}{2k}=\frac{z-3}{2}$ and $\frac{x-1}{3k}=\frac{y-5}{1}=\frac{z-6}{-5}$ are perpendicular to each other,then $k$ is

The equation of the line passing through the points $(3, 2, 4)$ and $(4, 5, 2)$ is

Let the image of the point $P(0, -5, 0)$ in the line $\frac{x-1}{2} = \frac{y}{1} = \frac{z+1}{-2}$ be the point $R$ and the image of the point $Q(0, -1/2, 0)$ in the line $\frac{x-1}{-1} = \frac{y+9}{4} = \frac{z+1}{1}$ be the point $S$. Then the square of the area of the parallelogram $PQRS$ is . . . . . . .

Find the acute angle between the line joining the points $(2, 1, -3)$ and $(-3, 1, 7)$ and the line parallel to $\frac{x - 1}{3} = \frac{y}{4} = \frac{z + 3}{5}$ passing through the point $(-1, 0, 4)$.