The angle between the two lines $\frac{x-3}{1}=\frac{y-2}{2}=\frac{z+4}{2}$ and $\frac{x-5}{3}=\frac{y+2}{2}=\frac{z}{6}$ is $\qquad$ .

The distance of the point $(-2, 4, -5)$ from the line $\frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6}$ is

Let the shortest distance between the lines $L : \frac{x-5}{-2} = \frac{y-\lambda}{0} = \frac{z+\lambda}{1}, \lambda \geq 0$ and $L_1 : x+1 = y-1 = 4-z$ be $2\sqrt{6}$. If $(\alpha, \beta, \gamma)$ lies on $L$,then which of the following is $NOT$ possible?

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:

If the lines $\frac{1-x}{2}=\frac{7y+4}{2\lambda}=\frac{2z-5}{2}$ and $\frac{7-7x}{3\lambda}=\frac{y-1}{7}=\frac{6-z}{5}$ are at right angles,then the value of $\lambda$ is

The image of the point $ (1,6,3) $ in the line $ \frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3} $ is