If the lines $\frac{1-x}{3}=\frac{7y-14}{2\lambda}=\frac{z-3}{2}$ and $\frac{7-7x}{3\lambda}=\frac{y-5}{1}=\frac{6-z}{5}$ are at right angles,then $\lambda=$

The distance of the point $-\hat{i} + 2\hat{j} + 6\hat{k}$ from the straight line that passes through the point $2\hat{i} + 3\hat{j} - 4\hat{k}$ and is parallel to the vector $6\hat{i} + 3\hat{j} - 4\hat{k}$ is

Let $P$ and $Q$ be the points on the line $\frac{x+3}{8}=\frac{y-4}{2}=\frac{z+1}{2}$ which are at a distance of $6$ units from the point $R(1,2,3)$. If the centroid of the triangle $PQR$ is $(\alpha, \beta, \gamma)$,then $\alpha^2+\beta^2+\gamma^2$ is:

The equation of a line passing through a point $(4, -2, 3)$ and perpendicular to the $XZ$-plane is:

If the line joining $A(4,1,2)$ and $B(0, k, 1)$ is perpendicular to the line joining $C(-2,1,1)$ and $D(4,2,5)$,then the value of $k$ is equal to

The vector equation of the line joining the points $i - 2j + k$ and $-2j + 3k$ is