The distance of the point $$(1, 3, -7)$$ from the plane passing through the point $$(1, -1, -1)$$, having normal perpendicular to both the lines $$\frac{x-1}{1} = \frac{y+2}{-2} = \frac{z-4}{3}$$ and $$\frac{x-2}{2} = \frac{y+1}{-1} = \frac{z+7}{-1}$$, is:

We have two given lines written in symmetric form as $$\frac{x-1}{1}=\frac{y+2}{-2}=\frac{z-4}{3}$$ and $$\frac{x-2}{2}=\frac{y+1}{-1}=\frac{z+7}{-1}.$$

From each line we read its direction ratios:

For the first line the direction ratios are $$1,\;-2,\;3,$$ so its direction vector is $$\vec{a}=(1,\,-2,\;3).$$

For the second line the direction ratios are $$2,\;-1,\;-1,$$ so its direction vector is $$\vec{b}=(2,\,-1,\,-1).$$

The normal vector of the required plane must be perpendicular to both the lines, hence it must be perpendicular to both $$\vec{a}$$ and $$\vec{b}.$$ To obtain a vector perpendicular to two given vectors we take their cross product. Thus we compute $$\vec{n}=\vec{a}\times\vec{b}.$$

Writing the cross product in determinant form, we get

$$ \vec{n}= \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ 1 & -2 & 3\\ 2 & -1 & -1 \end{vmatrix} =\mathbf{i}\bigl[(-2)(-1)-3(-1)\bigr] -\mathbf{j}\bigl[1(-1)-3(2)\bigr] +\mathbf{k}\bigl[1(-1)-(-2)(2)\bigr]. $$

Simplifying each component:

$$ \mathbf{i}:\;(-2)(-1)=2,\; 3(-1)=-3,\; 2-(-3)=5; $$

$$ \mathbf{j}:\;1(-1)=-1,\;3(2)=6,\; -1-6=-7,\; -(-7)=+7; $$

$$ \mathbf{k}:\;1(-1)=-1,\;(-2)(2)=-4,\; -1-(-4)=-1+4=3. $$

So the normal vector is $$\vec{n}=(5,\;7,\;3).$$

Now the required plane passes through the point $$(1,-1,-1).$$ Using the point-normal form of a plane, $$A(x-x_0)+B(y-y_0)+C(z-z_0)=0,$$ where $$(A,B,C)$$ are the normal’s components and $$(x_0,y_0,z_0)$$ is the given point, we substitute:

$$ 5(x-1)+7(y+1)+3(z+1)=0. $$

Expanding term by term, we obtain

$$ 5x-5+7y+7+3z+3=0. $$

Combining the constants, $$-5+7+3=5,$$ so the simplified Cartesian equation of the plane is

$$ 5x+7y+3z+5=0. $$

We now need the perpendicular distance of the point $$(1,3,-7)$$ from this plane. For a point $$(x_1,y_1,z_1)$$ and a plane $$Ax+By+Cz+D=0,$$ the distance formula is

$$ \text{Distance}=\frac{|Ax_1+By_1+Cz_1+D|}{\sqrt{A^2+B^2+C^2}}. $$

Here $$A=5,\;B=7,\;C=3,\;D=5,\;(x_1,y_1,z_1)=(1,3,-7).$$ Substituting, the numerator becomes

$$ |5(1)+7(3)+3(-7)+5| =|5+21-21+5| =|10| =10. $$

The denominator is

$$ \sqrt{5^2+7^2+3^2} =\sqrt{25+49+9} =\sqrt{83}. $$

Hence the required distance is

$$ \frac{10}{\sqrt{83}}. $$

Therefore, the distance matches Option B.

Hence, the correct answer is Option B.