Two lines $L_1: x=5, \frac{y}{3-\alpha}=\frac{z}{-2}$ and $L_2: x=\alpha, \frac{y}{-1}=\frac{z}{2-\alpha}$ are coplanar. Then $\alpha$ can take value$(s)$.

The acute angle between the lines $x=-y, z=0$ and $x=0, z=0$ is

The image of the point $ (1,6,3) $ in the line $ \frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3} $ is

Let $L$ be the line parallel to the vector $\sqrt{2} \hat{i}-5 \hat{j}+3 \hat{k}$ and passing through the point $A$ given by $\hat{i}+2 \hat{j}-3 \hat{k}$. If the distance between $A$ and a point $P$ on the line $L$ is $18$ units, then the position vector of such a point $P$ is

If the line passing through the points $(-5, 1, 3)$ and $(1, 2, 0)$ is perpendicular to the line passing through the points $(x, 2, 1)$ and $(0, -4, 6)$,then $x = \dots$

If the lines $\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-1}{\lambda}$ and $\frac{x-2}{3}=\frac{y-3}{2}=\frac{z-2}{3}$ are coplanar,then $\sin ^{-1}(\sin \lambda)+\cos ^{-1}(\cos \lambda)=$