Question

Find the equation plane passing $(a,b,c)$ and parallel to the plane $\overrightarrow{r}\cdot (\hat{i}+\hat{j}+\hat{k})=2$.

Answer 1

Any plane to parallel to the plane $\overrightarrow{r}.(\hat{i}+\hat{j}+\hat{k})=2$ is for the form
$\overrightarrow{r}.(\hat{i}+\hat{j}+\hat{k})=\lambda .....\left(1\right)$
The plane passes through the point $(a,b,c)$. Therefore, the position vector $\overrightarrow{r}$ of this point is $a\hat{i}+b\hat{j}+c\hat{k}$.
Therefore, equation (1) becomes
$(a\hat{i}+b\hat{j}+c\hat{k}).(\hat{i}+\hat{j}+\hat{k})=\lambda$
$\Rightarrow$ $a+b+c=\lambda$
Substituting $\lambda =a+b+c$ in equation (1) we obtain
$\overrightarrow{r}.(\hat{i}+\hat{j}+\hat{k})=a+b+c......\left(2\right)$
This is the vector equation of the required plane.
Substituting $\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$ in equation (2), we obtain
$(x\hat{i}+y\hat{j}+z\hat{k}).(\hat{i}+\hat{j}+\hat{k})=a+b+c$
$\Rightarrow$ $x+y+z=a+b+c$