The equation of the plane through the intersection of the planes $\vec{r}.(2\hat{i} + 6\hat{j}) + 12 = 0 $ and $\vec{r}.(3\hat{i} - \hat{j} + 4\hat{k})= 0 $ and at a unit distance from the origin, is

$\vec{r}.(2\hat{i} + \hat{j} + 2\hat{k})+3= 0 $

The equation of the plane through the intersection of the given planes is

$\begin{Bmatrix} \vec{r}.(2\hat{i} + 6\hat{j}) + 12\end{Bmatrix}+λ  \begin{Bmatrix}\vec{r}.(3\hat{i} - \hat{j} + 4\hat{k})\end{Bmatrix}= 0 $

or $\vec{r}.\begin{Bmatrix}(2 + 3λ)\hat{i} + (6 - λ)\hat{j} + 4λ \hat{k} + 12 \end{Bmatrix}= 0 $ ........(i)

It is at a unit distance from the origin

$∴\begin{vmatrix} \frac{12}{}\end{vmatrix}=1 ⇒ λ= ±2$

Hence, the equations of the required planes are

$\vec{r}.(2\hat{i} + \hat{j} + 2\hat{k})+3= 0 $ and $\vec{r}.(\hat{i} -2 \hat{j} + 2\hat{k})-3= 0 $