Given,

Distance from origin to the plane $\mathrm{d}=6 \text{units}$

And Normal vector $\overrightarrow{N}=2\hat{\mathrm{i}}+6\hat{\mathrm{j}}-3\hat{\mathrm{k}}$

$\therefore$Unit normal vector $\hat{n}=\frac{\overrightarrow{N}}{\left|\overrightarrow{N}\right|}=\frac{2\hat{\mathrm{i}}+6\hat{\mathrm{j}}-3\hat{\mathrm{k}}}{\sqrt{2^2+6^2+3^2}}=\frac{2\hat{\mathrm{i}}+6\hat{\mathrm{j}}-3\hat{\mathrm{k}}}{7}$

Equation of plane in normal form is

$\overrightarrow{r}·\hat{n}=d$

$\overrightarrow{r}·\frac{2\hat{i}+6\hat{j}-3\hat{k}}{7}=6$

Put $\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$

$\Rightarrow \left(x\hat{i}+y\hat{j}+z\hat{k}\right)·\left(2\hat{i}+6\hat{j}-3\hat{k}\right)=42$

$\therefore 2x+6y-3z-42=0$.