Determine the direction cosines of the unit vector perpendicular to the plane $\vec{r}\cdot (2\hat{i}-6\hat{j}-3\hat{k})+1=0$ passing through the origin?

The equation of the plane is $\vec{r}\cdot (2\hat{i}-6\hat{j}-3\hat{k})=-1$.To bring it to the normal form $\vec{r}\cdot \hat{n}=d$ where $d>0$, multiply by $-1$:
$$\vec{r}\cdot (-2\hat{i}+6\hat{j}+3\hat{k})=1$$
Here, the normal vector $\vec{n}=-2\hat{i}+6\hat{j}+3\hat{k}$.The magnitude of $\vec{n}$ is $|\vec{n}|=\sqrt{(-2{)}^2+6^2+3^2}=\sqrt{4+36+9}=\sqrt{49}=7$.The unit normal vector $\hat{n}=\frac{\vec{n}}{|\vec{n}|}=-\frac{2}{7}\hat{i}+\frac{6}{7}\hat{j}+\frac{3}{7}\hat{k}$.The direction cosines are the components of the unit vector:
$$\left(-\frac{2}{7},\frac{6}{7},\frac{3}{7}\right)$$