If the lines $\frac{1-x}{2}=\frac{y-8}{\lambda}=\frac{z-5}{2}$ and $\frac{x-11}{5}=\frac{y-3}{3}=\frac{z-1}{1}$ are perpendicular,then $\lambda=$

Points $P$ and $Q$ are given by $\vec{OP} = \hat{i} - \hat{j} - \hat{k}$ and $\vec{OQ} = -\hat{i} + \hat{j} + \hat{k}$. $A$ line along the vector $\vec{a} = \hat{i} + \hat{j}$ passes through the point $P$ and another line along the vector $\vec{b} = \hat{j} - \hat{k}$ passes through the point $Q$. If a line along the vector $\vec{c} = \hat{i} - \hat{j} + \hat{k}$ intersects both the lines along the vectors $\vec{a}$ and $\vec{b}$ at $L$ and $M$ respectively,then $\vec{PM} =$

If two lines $L_1$ and $L_2$ in space are defined by $L_1 = \{ x = \sqrt{\lambda} y + (\sqrt{\lambda} - 1), z = (\sqrt{\lambda} - 1)y + \sqrt{\lambda} \}$ and $L_2 = \{ x = \sqrt{\mu} y + (1 - \sqrt{\mu}), z = (1 - \sqrt{\mu})y + \sqrt{\mu} \}$,then $L_1$ is perpendicular to $L_2$ for all non-negative reals $\lambda$ and $\mu$ such that:

Find the foot of the perpendicular from the point $(0,2,3)$ on the line $\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}$.

The coordinates of the point at which the $XZ$-plane divides the line segment joining $A(-2, 3, 4)$ and $B(1, 2, 3)$ are:

The angle between the straight lines $\frac{x - 2}{2} = \frac{y - 1}{5} = \frac{z + 3}{-3}$ and $\frac{x + 1}{-1} = \frac{y - 4}{8} = \frac{z - 5}{4}$ is