Find the equation of the plane through the intersection of the planes $\vec{r} \cdot(\hat{i}+3 \hat{j})-6=0$ and $\vec{r} \cdot(3 \hat{i}-\hat{j}-4 \hat{k})=0,$ whose perpendicular distance from the origin is unity.

(A) The equation of the plane passing through the intersection of two planes $\vec{r} \cdot \vec{n}_{1} = d_{1}$ and $\vec{r} \cdot \vec{n}_{2} = d_{2}$ is given by $\vec{r} \cdot (\vec{n}_{1} + \lambda \vec{n}_{2}) = d_{1} + \lambda d_{2}$.
Given planes are $\vec{r} \cdot (\hat{i} + 3\hat{j}) = 6$ and $\vec{r} \cdot (3\hat{i} - \hat{j} - 4\hat{k}) = 0$.
So,the equation of the required plane is $\vec{r} \cdot [(\hat{i} + 3\hat{j}) + \lambda(3\hat{i} - \hat{j} - 4\hat{k})] = 6 + \lambda(0)$.
$\Rightarrow \vec{r} \cdot [(1 + 3\lambda)\hat{i} + (3 - \lambda)\hat{j} - 4\lambda\hat{k}] = 6$.
The perpendicular distance from the origin to the plane $\vec{r} \cdot \vec{n} = d$ is $\frac{|d|}{|\vec{n}|}$.
Here,$\frac{|6|}{\sqrt{(1 + 3\lambda)^{2} + (3 - \lambda)^{2} + (-4\lambda)^{2}}} = 1$.
$\Rightarrow 36 = (1 + 9\lambda^{2} + 6\lambda) + (9 + \lambda^{2} - 6\lambda) + 16\lambda^{2}$.
$\Rightarrow 36 = 26\lambda^{2} + 10$.
$\Rightarrow 26\lambda^{2} = 26 \Rightarrow \lambda^{2} = 1 \Rightarrow \lambda = \pm 1$.
For $\lambda = 1$,the equation is $\vec{r} \cdot (4\hat{i} + 2\hat{j} - 4\hat{k}) = 6 \Rightarrow 4x + 2y - 4z = 6 \Rightarrow 2x + y - 2z = 3$.
For $\lambda = -1$,the equation is $\vec{r} \cdot (-2\hat{i} + 4\hat{j} + 4\hat{k}) = 6 \Rightarrow -2x + 4y + 4z = 6 \Rightarrow -x + 2y + 2z = 3$.