The coordinates of the foot of the perpendicular from the point $$(1, -2, 1)$$ on the plane containing the lines $$\frac{x+1}{6} = \frac{y-1}{7} = \frac{z-3}{8}$$ and $$\frac{x-1}{3} = \frac{y-2}{5} = \frac{z-3}{7}$$, is:

We have two lines:

$$\dfrac{x+1}{6}=\dfrac{y-1}{7}=\dfrac{z-3}{8} \qquad\text{and}\qquad \dfrac{x-1}{3}=\dfrac{y-2}{5}=\dfrac{z-3}{7}.$$

Writing the first line in parametric form, we put the common ratio equal to a parameter $$t$$, obtaining

$$x=-1+6t,\; y=1+7t,\; z=3+8t.$$

Hence a point on this line is $$P_1(-1,1,3)$$ (take $$t=0$$) and its direction ratios are $$\langle 6,7,8\rangle.$$

For the second line we write the common ratio as a parameter $$s$$, giving

$$x=1+3s,\; y=2+5s,\; z=3+7s.$$

So a point on this line is $$P_2(1,2,3)$$ (take $$s=0$$) and its direction ratios are $$\langle 3,5,7\rangle.$$

The plane that contains both lines must contain the vectors along the lines. Therefore a normal vector to the plane can be obtained by taking the cross-product of these two direction vectors.

Let $$\vec a=\langle 6,7,8\rangle$$ and $$\vec b=\langle 3,5,7\rangle.$$ Then

$$\vec n=\vec a\times\vec b =\begin{vmatrix} \mathbf i & \mathbf j & \mathbf k\\ 6 & 7 & 8\\ 3 & 5 & 7 \end{vmatrix} =\mathbf i(7\cdot7-8\cdot5)-\mathbf j(6\cdot7-8\cdot3)+\mathbf k(6\cdot5-7\cdot3).$$

Calculating each component:

$$7\cdot7-8\cdot5=49-40=9,$$

$$6\cdot7-8\cdot3=42-24=18,$$

$$6\cdot5-7\cdot3=30-21=9.$$

So $$\vec n=\langle 9,-18,9\rangle=9\langle 1,-2,1\rangle.$$

Because any non-zero scalar multiple of a normal works, we choose the simpler normal $$\langle 1,-2,1\rangle.$$

Using this normal and the point $$P_1(-1,1,3)$$ in the plane formula $$\vec n\cdot(\vec r-\vec r_0)=0,$$ we write

$$1(x+1)-2(y-1)+1(z-3)=0.$$

Simplifying term by term,

$$x+1-2y+2+z-3=0 \;\;\Longrightarrow\;\; x-2y+z=0.$$

Thus the required plane is

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

Now we are asked for the foot of the perpendicular from the point $$A(1,-2,1)$$ to this plane. For the plane

$$ax+by+cz+d=0,$$

the coordinates $$\bigl(x',y',z'\bigr)$$ of the foot of the perpendicular from $$\bigl(x_0,y_0,z_0\bigr)$$ are given by the formula

$$x'=x_0-\dfrac{a(ax_0+by_0+cz_0+d)}{a^2+b^2+c^2},$$ $$y'=y_0-\dfrac{b(ax_0+by_0+cz_0+d)}{a^2+b^2+c^2},$$ $$z'=z_0-\dfrac{c(ax_0+by_0+cz_0+d)}{a^2+b^2+c^2}.$$

In our plane we have $$a=1,\; b=-2,\; c=1,\; d=0.$$ For the point $$A(1,-2,1)$$ we first compute

$$S=ax_0+by_0+cz_0+d =1\cdot1+(-2)\cdot(-2)+1\cdot1+0 =1+4+1=6.$$

The denominator is

$$a^2+b^2+c^2=1^2+(-2)^2+1^2=1+4+1=6.$$

Therefore

$$x'=1-\frac{1\cdot6}{6}=1-1=0,$$

$$y'=-2-\frac{(-2)\cdot6}{6}=-2-(-2)=0,$$

$$z'=1-\frac{1\cdot6}{6}=1-1=0.$$

So the foot of the perpendicular is the point $$\bigl(0,0,0\bigr).$$

Hence, the correct answer is Option C.