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:

If $(a, b, c)$ are the direction ratios of a line joining the points $(4, 3, -5)$ and $(-2, 1, -8)$,then the point $P(a, 3b, 2c)$ lies on the plane:

Let $L_1$ (respectively $L_2$) be the line passing through $2 \hat{i}-\hat{k}$ (respectively $2 \hat{i}+\hat{j}-3 \hat{k}$) and parallel to $3 \hat{i}-\hat{j}+2 \hat{k}$ (respectively $\hat{i}-2 \hat{j}+\hat{k}$). Then the shortest distance between the lines $L_1$ and $L_2$ is equal to

Two lines $\frac{x - 3}{1} = \frac{y + 1}{3} = \frac{z - 6}{-1}$ and $\frac{x + 5}{7} = \frac{y - 2}{-6} = \frac{z - 3}{4}$ intersect at the point $R$. The reflection of $R$ in the $xy$-plane has coordinates

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 $\vec{PM} =$

Lines $\frac{1-x}{3}=\frac{y-2}{1}=\frac{z-1}{2}$ and $\frac{x-2}{p}=\frac{y-1}{2}=\frac{z-2}{1}$ are mutually perpendicular to each other,then $p=$ . . . . . . .