1. Write the given lines in parametric form

The two given lines are

$\frac{x-7}{3}=\frac{y-1}{-1}=\frac{z+2}{1}$ and $\frac{x}{2}=\frac{y-7}{3}=\frac{z}{1}.$

Let the first line be parameterized by $t$:

$x=7+3t,\quad y=1-t,\quad z=-2+t.$

So a general point on the first line is

$A(7+3t,\,1-t,\,-2+t).$

Let the second line be parameterized by $s$:

$x=2s,\quad y=7+3s,\quad z=s.$

So a general point on the second line is

$B(2s,\,7+3s,\,s).$

2. Use the condition on the required line

The line through $A$ and $B$ has direction ratios $(1,-4,2)$. Therefore,

$\overrightarrow{AB}=B-A$ must be parallel to $(1,-4,2)$.

Now,

$\overrightarrow{AB}=(2s-(7+3t),\,(7+3s)-(1-t),\,s-(-2+t)).$

So,

$\overrightarrow{AB}=(2s-7-3t,\,6+3s+t,\,s+2-t).$

Since this is parallel to $(1,-4,2)$, there exists some scalar $k$ such that

$2s-7-3t=k,$ $6+3s+t=-4k,$ $s+2-t=2k.$

3. Solve for $s,t,k$

From the first equation,

$k=2s-7-3t.$

From the third equation,

$s+2-t=2(2s-7-3t).$

Simplify:

$s+2-t=4s-14-6t$ $16=3s-5t. \quad \text{(1)}$

Now use the second equation:

$6+3s+t=-4(2s-7-3t)$ $6+3s+t=-8s+28+12t$ $11s-11t=22$ $s-t=2. \quad \text{(2)}$

From (2),

$s=t+2.$

Substitute into (1):

$16=3(t+2)-5t$ $16=3t+6-5t$ $16=6-2t$ $-2t=10$ $t=-5.$

Hence,

$s=t+2=-3.$

4. Find points $A$ and $B$

For $t=-5$,

$A=(7+3(-5),\,1-(-5),\,-2+(-5))=(-8,6,-7).$

For $s=-3$,

$B=(2(-3),\,7+3(-3),\,-3)=(-6,-2,-3).$

5. Compute $AB^2$

$\overrightarrow{AB}=B-A=(-6+8,\,-2-6,\,-3+7)=(2,-8,4).$

Therefore,

$AB^2=2^2+(-8)^2+4^2=4+64+16=84.$

6. Final answer

$\boxed{84}$

This matches the stored correct answer.