The distance of the point $(7,5,2)$ from the plane $3x+4y+z-8=0$ measured parallel to the line $\frac{x-1}{3}=\frac{y-2}{6}=\frac{z+1}{2}$ is:

Let $P$ be the plane $\sqrt{3} x+2 y+3 z=16$ and let $S=\left\{\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}: \alpha^2+\beta^2+\gamma^2=1 \text{ and the distance of } (\alpha, \beta, \gamma) \text{ from the plane } P \text{ is } \frac{7}{2}\right\}$. Let $\overrightarrow{u}, \overrightarrow{v}$ and $\overrightarrow{w}$ be three distinct vectors in $S$ such that $|\overrightarrow{u}-\overrightarrow{v}|=|\overrightarrow{v}-\overrightarrow{w}|=|\overrightarrow{w}-\overrightarrow{u}|$. Let $V$ be the volume of the parallelepiped determined by vectors $\overrightarrow{u}, \overrightarrow{v}$ and $\overrightarrow{w}$. Then the value of $\frac{80}{\sqrt{3}} V$ is

The length of the projection of the line segment joining the points $(5,-1,4)$ and $(4,-1,3)$ on the plane $x+y+z=7$ is

If the angle $\theta$ between the line $\frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2}$ and the plane $2x-y+\sqrt{\lambda}z+4=0$ is such that $\sin \theta=\frac{1}{3}$,then the value of $\lambda=$

The ratio in which the plane $x - 2y + 3z = 17$ divides the line segment joining the points $A(-2, 4, 7)$ and $B(3, -5, 8)$ is:

$A$ plane $P$ contains the line of intersection of the planes $\vec{r} \cdot (\hat{i}+\hat{j}+\hat{k}) = 6$ and $\vec{r} \cdot (2\hat{i}+3\hat{j}+4\hat{k}) = -5$. If $P$ passes through the point $(0, 2, -2)$,then the square of the distance of the point $(12, 12, 18)$ from the plane $P$ is