Find the vector and cartesian equations of a plane which is at a distance of 18 units from the origin and which is normal to the vector $2 \hat{i} + 7 \hat{j} + 8 \hat{k}$.
Vector equation of a plane at a distance $p$ from the origin and normal to unit vector $\hat{n}$ is
$\overrightarrow{r} \cdot \hat{n} = p$ $\;\;\; \cdots \; (1)$
Here, $p = 18$ units;
$\overrightarrow{n} = 2 \hat{i} + 7 \hat{j} + 8 \hat{k}$
Now,
Unit vector $\hat{n} = \dfrac{\overrightarrow{n}}{\left|\overrightarrow{n}\right|}$
i.e. $\;$ $\hat{n} = \dfrac{2 \hat{i} + 7 \hat{j} + 8 \hat{k}}{\sqrt{\left(2\right)^2 + \left(7\right)^2 + \left(8\right)^2}} = \dfrac{2 \hat{i} + 7 \hat{j} + 8 \hat{k}}{\sqrt{117}} = \dfrac{2 \hat{i} + 7 \hat{j} + 8 \hat{k}}{3 \sqrt{13}}$
$\therefore$ $\;$ The required vector equation of the plane is [from equation $(1)$]
$\overrightarrow{r} \cdot \left(\dfrac{2 \hat{i} + 7 \hat{j} + 8 \hat{k}}{3\sqrt{13}}\right) = 18$ $\;\;\; \cdots \; (2)$
Putting $\overrightarrow{r} = x \hat{i} + y \hat{j} + z \hat{k}$ $\;$ in equation $(2)$, the cartesian equation of the plane is
$\left(x \hat{i} + y \hat{j} + z \hat{k}\right) \cdot \left(2 \hat{i} + 7 \hat{j} + 8 \hat{k}\right) = 54 \sqrt{13}$
i.e. $\;$ $2x + 7 y + 8 z = 54 \sqrt{13}$