$ABC$ is a triangle in a plane with vertices $A(2, 3, 5)$,$B(-1, 3, 2)$,and $C(\lambda, 5, \mu)$. If the median through $A$ is equally inclined to the coordinate axes,then the value of $\lambda + \mu$ 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 point $(-1, \alpha, \beta)$ lie on the line of the shortest distance between the lines $\frac{x+2}{-3}=\frac{y-2}{4}=\frac{z-5}{2}$ and $\frac{x+2}{-1}=\frac{y+6}{2}=\frac{z-1}{0}$. Then $(\alpha-\beta)^2$ is equal to ....................

The cosine of the angle included between the lines $\overline{r}=(2 \hat{\imath}+\hat{\jmath}-2 \hat{k})+\lambda(\hat{\imath}-2 \hat{\jmath}-2 \hat{k})$ and $\overline{r}=(\hat{\imath}+\hat{\jmath}+3 \hat{k})+\mu(3 \hat{\imath}+2 \hat{\jmath}-6 \hat{k})$ where $\lambda, \mu \in R$ is

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 $\sin ^{-1}(\sin \lambda)+\cos ^{-1}(\cos \lambda)=$

Let the image of the point $P(1, 2, 3)$ in the line $L : \frac{x-6}{3} = \frac{y-1}{2} = \frac{z-2}{3}$ be $Q$. Let $R(\alpha, \beta, \gamma)$ be a point that divides the line segment $PQ$ internally in the ratio $1:3$. Then the value of $22(\alpha+\beta+\gamma)$ is equal to