K. C. Sinha and Abhishek Chandra Solutions for Exercise 2: EXERCISE

The distance of the point $(2, 1, -1)$ from the plane $x-2y+4z=9$ is

The equation of the plane passing through the intersection of the planes $x+2y+3z+4=0$ and $4x+3y+2z+1=0$ and the origin is

If $p$ is the length of perpendicular from the origin onto the plane whose intercepts on the axes are $a, b, c$ then:

Consider the points $\mathrm{P}(p, 0, 0), \mathrm{Q}(0, q, 0)$ and $\mathrm{R}(0, 0, r)$, where $pqr\ne 0$, then the equation of the plane $PQR$ is:

The planes $x=0$ and $y=0$:

The plane $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$:

The distance of the plane $\overrightarrow{r}·\left(\frac{2}{7}\hat{i}+\frac{3}{7}\hat{j}-\frac{6}{7}\hat{k}\right)=1$ from the origin is:

The planes $\overrightarrow{r}·(2\hat{i}+3\hat{j}-6\hat{k})=7$ and $\overrightarrow{r}·\left(\frac{-2}{7}\hat{i}-\frac{3}{7}\hat{j}+\frac{6}{7}\hat{k}\right)=0$ are: