If the lines $\frac{2x-4}{\lambda} = \frac{y-1}{2} = \frac{z-3}{1}$ and $\frac{x-1}{1} = \frac{3y-1}{\lambda} = \frac{z-2}{1}$ are perpendicular to each other,then $\lambda=$

Find the position vector of the image of the point with position vector $\vec{P} = 2\hat{i} + \hat{j} + 3\hat{k}$ in the line whose vector equation is $\vec{r} = \hat{j} - 2\hat{k} + \lambda(\hat{i} + \hat{j} - \hat{k})$.

If $d_1$ is the shortest distance between the lines $x+1=2y=-12z$ and $x=y+2=6z-6$,and $d_2$ is the shortest distance between the lines $\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5}$ and $\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}$,then the value of $\frac{32 \sqrt{3} d_1}{d_2}$ is:

Show that the line joining the origin to the point $(2,1,1)$ is perpendicular to the line determined by the points $(3,5,-1)$ and $(4,3,-1).$

Let the line $L$ intersect the lines $x-2=-y=z-1$ and $2(x+1)=2(y-1)=z+1$,and be parallel to the line $\frac{x-2}{3}=\frac{y-1}{1}=\frac{z-2}{2}$. Then which of the following points lies on $L$?

The coordinates of the point where the line joining the points $(3, 5, -7)$ and $(-2, 1, 8)$ is intersected by the $yz$-plane are given by: