The length and foot of the perpendicular from the point $(7, 14, 5)$ to the plane $2x + 4y - z = 2$ are

If the angle between the line $x = \frac{y - 1}{2} = \frac{z - 3}{\lambda}$ and the plane $x + 2y + 3z = 4$ is $\cos^{-1}\left(\sqrt{\frac{5}{14}}\right)$,then $\lambda = \dots$

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 $\lambda =$

Let $P$ be a plane passing through the points $(1,0,1), (1,-2,1)$ and $(0,1,-2)$. Let a vector $\vec{a} = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}$ be such that $\vec{a}$ is parallel to the plane $P$,perpendicular to $(\hat{i} + 2 \hat{j} + 3 \hat{k})$ and $\vec{a} \cdot (\hat{i} + \hat{j} + 2 \hat{k}) = 2$. Then $(\alpha - \beta + \gamma)^2$ equals:

The angle between the line $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$ and the plane $3x + 2y - 3z = 4$ is ......... $^o$

If $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-7}{2}$ lies in the plane $ax+by+z=7$, then $a+b=$