The distance of line $$3y - 2z - 1 = 0 = 3x - z + 4$$ from the point $$(2, -1, 6)$$ is:

We are given the line as the common intersection of two planes

$$3y - 2z - 1 = 0 \qquad\text{and}\qquad 3x - z + 4 = 0.$$

For such a line we first find a direction vector by taking the cross product of the normals of the two planes.

The normal to the first plane is $$\vec n_1 = (0,\,3,\,-2)$$ and the normal to the second plane is $$\vec n_2 = (3,\,0,\,-1).$$

Taking the cross product, we have

$$ \vec d=\vec n_1 \times \vec n_2 =\begin{vmatrix} \hat i & \hat j & \hat k\\ 0 & 3 & -2\\ 3 & 0 & -1 \end{vmatrix} = \hat i\,(3\cdot(-1)-(-2\cdot0)) -\hat j\,(0\cdot(-1)-(-2\cdot3)) +\hat k\,(0\cdot0-3\cdot3) =(-3,\,-6,\,-9). $$

Dividing by $$-3$$ for simplicity, the direction vector is

$$\vec d=(1,\,2,\,3).$$

Next, we need one point on the line. Putting $$x=0$$ in the second plane gives

$$3(0)-z+4=0\;\Longrightarrow\;z=4.$$

Substituting $$z=4$$ in the first plane,

$$3y-2(4)-1=0\;\Longrightarrow\;3y-9=0\;\Longrightarrow\;y=3.$$

So a convenient point on the line is

$$A(0,\,3,\,4).$$

The given external point is

$$P(2,\,-1,\,6).$$

We form the vector $$\overrightarrow{AP}$$ from point $$A$$ to point $$P$$:

$$\overrightarrow{AP}=P-A=(2-0,\,-1-3,\,6-4)=(2,\,-4,\,2).$$

To find the distance of a point from a line we use the formula

$$\text{Distance}=\dfrac{\lvert\overrightarrow{AP}\times\vec d\rvert}{\lvert\vec d\rvert}.$$

First compute the cross product $$\overrightarrow{AP}\times\vec d$$:

$$ \overrightarrow{AP}\times\vec d =\begin{vmatrix} \hat i & \hat j & \hat k\\ 2 & -4 & 2\\ 1 & 2 & 3 \end{vmatrix} = \hat i\,((-4)(3)-(2)(2)) -\hat j\,(2\cdot3-2\cdot1) +\hat k\,(2\cdot2-(-4)\cdot1) =(-16,\,-4,\,8). $$

The magnitude of this vector is

$$ \lvert\overrightarrow{AP}\times\vec d\rvert =\sqrt{(-16)^2+(-4)^2+8^2} =\sqrt{256+16+64} =\sqrt{336} =4\sqrt{21}. $$

The magnitude of the direction vector is

$$ \lvert\vec d\rvert=\sqrt{1^2+2^2+3^2} =\sqrt{1+4+9} =\sqrt{14}. $$

Hence the required distance is

$$ \text{Distance} =\dfrac{4\sqrt{21}}{\sqrt{14}} =4\sqrt{\frac{21}{14}} =4\sqrt{\frac{3}{2}} =4\cdot\frac{\sqrt{6}}{2} =2\sqrt{6}. $$

Hence, the correct answer is Option B.