The equation of the plane passing through the line of intersection of the planes $x + y + z = 1$ and $2x + 3y - z + 4 = 0$ and parallel to the $x$-axis is:

The distance of the point $(1, 1, 9)$ from the point of intersection of the line $\frac{x-3}{1} = \frac{y-4}{2} = \frac{z-5}{2}$ and the plane $x+y+z=17$ 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 point where the line $\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+3}{4}$ meets the plane $2x+4y-z=1$ is:

Find the distance of the point $2\hat{i} + \hat{j} - \hat{k}$ from the plane $\vec{r} \cdot (\hat{i} - 2\hat{j} + 4\hat{k}) = 9$.

The direction ratios of the normal to the plane passing through the points $(1, 2, -3)$,$(-1, -2, 1)$ and parallel to $\frac{x-2}{2}=\frac{y+1}{3}=\frac{z}{4}$ are: