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

Given two planes,

$\vec{r}.\left ( \hat{i}+3\hat{j} \right )-6=0\\ \vec{r}.\left(3\hat{i}-\hat{j}-4\hat{k} \right )=0$

Also given, the perpendicular distance of the plane from the origin, = 1.

We must find the equation for this plane.

We know,

$\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$

Simplifying the planes,

$\vec{r}.\left ( \hat{i}+3\hat{j} \right )-6=0\\ \Rightarrow \left ( x\hat{i}+y\hat{j}+z\hat{k} \right ).\left ( \hat{i}+3\hat{j} \right )-6=0\\ \Rightarrow x+3y-6=0........(i)$

Also, for

$\vec{r}.\left(3\hat{i}-\hat{j}-4\hat{k} \right )=0\\ \Rightarrow \left ( x\hat{i}+y\hat{j}+z\hat{k} \right ).\left(3\hat{i}-\hat{j}-4\hat{k} \right )=0\\ \Rightarrow 3x-y-4z=0$

The equation of a plane through the line of intersection of x +   3y - 6 = 0 and 3x - y - 4z = 0 can be given as
$(x + 3y - 6) + \lambda(3x - y - 4z) = 0\\ \Rightarrow x + 3y - 6 + 3\lambda x - \lambda y - 4\lambda z = 0\\ \Rightarrow x + 3\lambda x + 3y - \lambda y - 6 - 4\lambda z = 0\\ \Rightarrow (1 + 3\lambda)x + (3 - \lambda)y - 4\lambda z - 6 = 0 %u2026(iii)$

Also, we know, the perpendicular distance of a plane, ax + by + cz + d = 0 from the origin, let’s say P, is given by

$P=\left | \frac{d}{\sqrt{a^{2}+b^{2}+c^{2}}} \right |$

Similarly, the perpendicular distance of the plane in equation (iii) from the origin (=1 according to the question) is:

$1=\left | \frac{-6}{\sqrt{\left ( 1+3 \lambda \right )^{2}+\left ( 3-\lambda \right )^{2}+\left ( -4\lambda \right )^{2}}} \right |\\ \\ \Rightarrow \sqrt{\left ( 1+3 \lambda \right )^{2}+\left ( 3-\lambda \right )^{2}+\left ( -4\lambda \right )^{2}}=6$

Taking the square of both sides,

$\Rightarrow \left (\sqrt{\left ( 1+3 \lambda \right )^{2}+\left ( 3-\lambda \right )^{2}+\left ( -4\lambda \right )^{2}} \right )^{2}=6^{2}$
$\Rightarrow (1 + 3\lambda)^{2} + (3 - \lambda)^{2} + (-4\lambda)^{2} = 36\\ \Rightarrow 1 + (3\lambda)^{2} + 2(1)(3\lambda) + (3)^{2} + \lambda^2 - 2(3)(\lambda) + 16\lambda^{2} = 36\\ \Rightarrow 1 + 9\lambda^{2} + 6\lambda + 9 + \lambda^{2} - 6\lambda + 16\lambda^{2} = 36\\ \Rightarrow 9\lambda^{2} + 16\lambda^{2} + \lambda^{2} + 6\lambda - 6\lambda = 36 - 1 - 9\\ \Rightarrow 26\lambda^{2} + 0 = 26\\ => \lambda^{2} = 26/26\\ => \lambda^{2} = 1\\ => \lambda = \pm 1$

First, we substitute $\lambda=1$ in eq (iii) to find the plane equation
$(1 + 3\lambda)x + (3 - \lambda)y - 4\lambda z - 6 = `0\\ \Rightarrow (1 + 3(1))x - (3 - 1)y - 4(1)z - 6 = 0\\ \Rightarrow 4x - 2y - 4z- 6 = 0$

Now, we substitute λ= -1 in eq. (iii) to find the plane equation
$(1 + 3\lambda)x + (3 - \lambda)y - 4 \lambda z - 6 = 0\\ \Rightarrow (1 + 3(-1))x + (3 - (-1))y - 4(-1)z - 6 = 0\\ \Rightarrow (1 - 3)x + (3 + 1)y + 4z - 6 = 0\\ \Rightarrow -2x + 4y + 4z - 6 = 0$

Therefore, the equation of the required plane is -2x + 4y + 4z – 6 = 0 and 4x – 2y – 4z – 6 = 0.