If the lines given by $\frac{x - 1}{2 \lambda} = \frac{y - 1}{-5} = \frac{z - 1}{2}$ and $\frac{x + 2}{\lambda} = \frac{y + 3}{\lambda} = \frac{z + 5}{1}$ are parallel,then the value of $\lambda$ is

Find the coordinates of the foot of the perpendicular from the point $(2, 4, 1)$ to the line $\vec{r} = (-5, -3, 6) + k(1, 4, -9)$,where $k \in R$.

The coordinates of the point where the line passing through $A(3, 4, 1)$ and $B(5, 1, 6)$ crosses the $xy$-plane are

Let $L$ be a line passing through a point $A$ and parallel to the vector $2 \hat{i}+\hat{j}-2 \hat{k}$. Let $-7 \hat{i}-5 \hat{j}+11 \hat{k}$ be the position vector of a point $P$ on $L$ such that $|\overline{AP}|=12$. Then the position vector of $A$ can be

Let a line pass through two distinct points $P(-2, -1, 3)$ and $Q$,and be parallel to the vector $3\hat{i} + 2\hat{j} + 2\hat{k}$. If the distance of the point $Q$ from the point $R(1, 3, 3)$ is $5$,then the square of the area of $\triangle PQR$ is equal to:

$A$ line passing through the points $P(1, -1, 2)$ and $Q(2, 0, 1)$ meets the $XY$-plane and the $YZ$-plane at points $A$ and $B$ respectively. The distance $AB$ is equal to...