Equation of a plane at a distance $$\sqrt{\frac{2}{21}}$$ units from the origin, which contains the line of intersection of the planes $$x - y - z - 1 = 0$$ and $$2x + y - 3z + 4 = 0$$, is

We are asked to find that member of the family of planes which  (i)  passes through the line of intersection of the two given planes and  (ii)  is at the specified distance $$\sqrt{\dfrac{2}{21}}$$ from the origin.

First we recall that the equation of the family of planes containing the intersection of the two planes $$x - y - z - 1 = 0 \quad\text{and}\quad 2x + y - 3z + 4 = 0$$ is written by adding an arbitrary scalar multiple $$\lambda$$ of the second equation to the first. Thus we have

$$\bigl(x - y - z - 1\bigr) + \lambda\bigl(2x + y - 3z + 4\bigr)=0.$$

Simplifying term by term we obtain

$$\begin{aligned} (1+2\lambda)\,x \;+\; (-1+\lambda)\,y \;+\; (-1-3\lambda)\,z \;+\; (-1+4\lambda) &= 0,\\[4pt] \text{or}\qquad A x + B y + C z + D &= 0, \end{aligned}$$

where

$$A = 1+2\lambda$$, $$\qquad B = -1+\lambda$$, $$\qquad C = -1-3\lambda$$, $$\qquad D = -1+4\lambda.$$

Next we use the distance formula. For a plane $$Ax + By + Cz + D = 0$$, the perpendicular distance of the origin $$(0,0,0)$$ from the plane is

Distance $$\;=\; \frac{|D|}{\sqrt{A^{2}+B^{2}+C^{2}}}.$$

We are told that this distance equals $$\sqrt{\dfrac{2}{21}}$$. Therefore

$$\frac{|D|}{\sqrt{A^{2}+B^{2}+C^{2}}} \;=\; \sqrt{\dfrac{2}{21}}.$$

Because the distance is non-negative, we square both sides to remove the absolute value without sign ambiguity and obtain

$$\frac{D^{2}}{A^{2}+B^{2}+C^{2}} \;=\; \frac{2}{21},$$ $$\Longrightarrow\; 21D^{2} \;=\; 2\bigl(A^{2}+B^{2}+C^{2}\bigr).$$

We now compute the needed squares.

$$\begin{aligned} A^{2} &= (1+2\lambda)^{2} = 1 + 4\lambda + 4\lambda^{2},\\[4pt] B^{2} &= (-1+\lambda)^{2} = 1 - 2\lambda + \lambda^{2},\\[4pt] C^{2} &= (-1-3\lambda)^{2} = 1 + 6\lambda + 9\lambda^{2}. \end{aligned}$$

Adding these three results term by term gives

$$A^{2}+B^{2}+C^{2} = 3 + 8\lambda + 14\lambda^{2}.$$

Next we square $$D = -1 + 4\lambda$$ to get

$$D^{2} = (-1+4\lambda)^{2} = 1 - 8\lambda + 16\lambda^{2}.$$

Substituting $$A^{2}+B^{2}+C^{2}$$ and $$D^{2}$$ into the equation $$21D^{2} = 2(A^{2}+B^{2}+C^{2})$$ we have

$$21\bigl(1 - 8\lambda + 16\lambda^{2}\bigr) \;=\; 2\bigl(3 + 8\lambda + 14\lambda^{2}\bigr).$$

Expanding both sides gives

$$21 - 168\lambda + 336\lambda^{2} \;=\; 6 + 16\lambda + 28\lambda^{2}.$$

Moving every term to the left we find

$$21 - 168\lambda + 336\lambda^{2} - 6 - 16\lambda - 28\lambda^{2} = 0,$$ $$\Longrightarrow\; 15 - 184\lambda + 308\lambda^{2} = 0.$$

This is a quadratic equation in $$\lambda$$:

$$308\lambda^{2} - 184\lambda + 15 = 0.$$

We compute its discriminant:

$$\Delta = (-184)^{2} - 4\cdot 308 \cdot 15 = 33856 - 18480 = 15376.$$

Because $$15376 = 16 \times 961 = (4\times 31)^{2},$$ we have $$\sqrt{\Delta} = 124.$$ Therefore

$$\lambda \;=\; \frac{184 \pm 124}{2\cdot 308} = \frac{184 \pm 124}{616}.$$

Hence the two possible values of $$\lambda$$ are

$$\lambda_{1} = \frac{184 + 124}{616} = \frac{308}{616} = \frac{1}{2}, \qquad \lambda_{2} = \frac{184 - 124}{616} = \frac{60}{616} = \frac{15}{154}.$$

We now write the equation of the plane for each value.

For $$\lambda = \dfrac12$$:

$$\begin{aligned} A &= 1 + 2\!\left(\tfrac12\right) = 1 + 1 = 2,\\ B &= -1 + \tfrac12 = -\tfrac12,\\ C &= -1 - 3\!\left(\tfrac12\right) = -1 - \tfrac32 = -\tfrac52,\\ D &= -1 + 4\!\left(\tfrac12\right) = -1 + 2 = 1. \end{aligned}$$

The corresponding plane is therefore

$$2x - \tfrac12\,y - \tfrac52\,z + 1 = 0.$$

To clear the fractions we multiply the entire equation by 2:

$$4x - y - 5z + 2 = 0.$$

For $$\lambda = \dfrac{15}{154}$$ we would obtain another plane, but that plane does not match any of the given options, whereas the plane obtained above exactly coincides with Option 4.

Thus the required plane is

$$4x - y - 5z + 2 = 0.$$

Hence, the correct answer is Option 4.