If the line $\frac{x+1}{2}=\frac{y-m}{3}=\frac{z-4}{6}$ lies in the plane $3x-14y+6z+49=0$,then the value of $m$ is

$A$ plane $\pi$ is passing through the points $A(1, -2, 3)$ and $B(6, 4, 5)$. If the plane $\pi$ is perpendicular to the plane $3x - y + z = 2$,then the perpendicular distance from $(0, 0, 0)$ to the plane $\pi$ is

Let $P$ be the plane passing through the line $\frac{x-1}{1}=\frac{y-2}{-3}=\frac{z+5}{7}$ and the point $(2,4,-3)$. If the image of the point $(-1,3,4)$ in the plane $P$ is $(\alpha, \beta, \gamma)$,then $\alpha+\beta+\gamma$ is equal to

The distance of the point $O(\vec{0})$ from the plane $\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=5$ measured parallel to the vector $2 \hat{i}+3 \hat{j}-6 \hat{k}$ is:

$A$ line $L$ passes through the points $\hat{i}+2 \hat{j}+\hat{k}$ and $-2 \hat{i}+3 \hat{k}$. $A$ plane $P$ passes through the origin and the points $4 \hat{k}, 2 \hat{i}+\hat{j}$. The point where the line $L$ meets the plane $P$ 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 $\lambda =$