Concept:
The distance of the origin $\left(0,0,0\right)$ from the plane $\mathit{a}\mathit{x}+\mathit{b}\mathit{y}+\mathit{c}\mathit{z}+\mathit{d}=0$ is given by $\left|\frac{\left(\mathit{a}\right)\left(0\right)+\left(\mathit{b}\right)\left(0\right)+\left(\mathit{c}\right)\left(0\right)+\mathit{d}}{\sqrt{{\mathit{a}}^2+{\mathit{b}}^2+{\mathit{c}}^2}}\right|$
Calculation:
We know that the distance of the origin $\left(0,0,0\right)$ from the plane $\mathit{a}\mathit{x}+\mathit{b}\mathit{y}+\mathit{c}\mathit{z}+\mathit{d}=0$ is given by $\left|\frac{\left(\mathrm{a}\right)\left(0\right)+\left(\mathrm{b}\right)\left(0\right)+\left(\mathrm{c}\right)\left(0\right)+\mathrm{d}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}}\right|$
$\Rightarrow$ The distance of the origin from the plane $2\mathit{x}+6\mathit{y}-3\mathit{z}+7=0$
$=\left|\frac{\left(2\right)\left(0\right)+\left(6\right)\left(0\right)+\left(-3\right)\left(0\right)+7}{\sqrt{2^2+6^2+{\left(-3\right)}^2}}\right|$
$=\left|\frac{7}{\sqrt{49}}\right|$
$=1$