Let the plane$\overrightarrow{r}.\overrightarrow{n_1}=d_1$ be $\overrightarrow{r}.(\overrightarrow{i}+3\overrightarrow{j})=6 \rightarrow d_1=\frac{6}{\sqrt {10}}$
Let the other plane be $\overrightarrow{r}.\overrightarrow{n_2}=d_2$ be $\overrightarrow{r}.(\overrightarrow{3i}-\overrightarrow{j}-4\overrightarrow{k})=0 \rightarrow d_2=0$
$\textbf{Step 2}$:
Therefore $\overrightarrow{r}.(n_1+\lambda \overrightarrow{n_2})=d_1+\lambda d_2$ represents the plane passing through the intersection of the above two planes.
Given $d_1+\lambda d_2=1$ $\Rightarrow \lambda =\frac{\sqrt {10}}{6}$
$\textbf{Step 3}$:
Therefore the required equation is
$\overrightarrow{r}.\bigg[(\overrightarrow{i}+3\overrightarrow{j})+\frac{\sqrt {10}}{6}(3\overrightarrow{i}-\overrightarrow{j}-4\overrightarrow{k})\bigg]=1$
$\overrightarrow{r}.\bigg[6\overrightarrow{i}+18\overrightarrow{j}+3\sqrt {10}\overrightarrow{i}-10\overrightarrow{j}-4 \sqrt {10} \overrightarrow{k}\bigg]=6$
$\Rightarrow \overrightarrow{r}.[\overrightarrow{i}(6+3 \sqrt {10})+\overrightarrow{j}(18-\sqrt {10})-4 \sqrt {10}\overrightarrow{k})=6$