Let $\frac{x-2}{3}=\frac{y+1}{-2}=\frac{z+3}{-1}$ lie on the plane $px-qy+z=5$,for some $p, q \in R$. The shortest distance of the plane from the origin is

The angle between the line with the direction ratios $(2, 5, 1)$ and the plane $8x + 2y - z = 4$ is

If the line $\bar{r}=(\hat{i}-2 \hat{j}+3 \hat{k})+\lambda(2 \hat{i}+\hat{j}+2 \hat{k})$ is parallel to the plane $\bar{r} \cdot(3 \hat{i}-2 \hat{j}-m \hat{k})=5$,then the value of $m$ is:

If ${L_1}$ is the line of intersection of the planes $2x - 2y + 3z - 2 = 0$ and $x - y + z + 1 = 0$,and ${L_2}$ is the line of intersection of the planes $x + 2y - z - 3 = 0$ and $3x - y + 2z - 1 = 0$,then the distance of the origin from the plane containing the lines ${L_1}$ and ${L_2}$ is:

The line passing through the points $(1, 1, -1)$ and $(3, -1, 0)$ makes an angle of $\operatorname{Tan}^{-1}\left(\frac{1}{\sqrt{8}}\right)$ with the plane $\sqrt{\lambda} x + 3y + 6z = 17$. Then $\lambda =$

Suppose the line $\frac{x-2}{\alpha}=\frac{y-2}{-5}=\frac{z+2}{2}$ lies on the plane $x+3y-2z+\beta=0$. Then $(\alpha+\beta)$ is equal to ... .