Let the three points be

$A(\alpha,10,13),\quad B(6,11,11),\quad C\left(\frac{9}{2},\beta,-8\right).$

Since the points are collinear, the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ must be parallel.

1. Compute the direction vectors

$\overrightarrow{AB}=B-A=(6-\alpha,\,1,\,-2)$

and

$\overrightarrow{AC}=C-A=\left(\frac{9}{2}-\alpha,\,\beta-10,\,-21\right).$

For collinearity,

$\overrightarrow{AC}=\lambda\,\overrightarrow{AB}$

for some scalar $\lambda$.

So,

$\left(\frac{9}{2}-\alpha,\,\beta-10,\,-21\right)=\lambda(6-\alpha,\,1,\,-2).$

2. Use the $z$-coordinates to find $\lambda$

From the third component,

$-21=\lambda(-2)$

$\lambda=\frac{21}{2}.$

3. Find $\alpha$

From the first component,

$\frac{9}{2}-\alpha=\frac{21}{2}(6-\alpha).$

Multiply by $2$:

$9-2\alpha=21(6-\alpha)$

$9-2\alpha=126-21\alpha$

$19\alpha=117$

$\alpha=\frac{117}{19}.$

4. Find $\beta$

From the second component,

$\beta-10=\frac{21}{2}(1)=\frac{21}{2}$

$\beta=10+\frac{21}{2}=\frac{41}{2}.$

5. Compute $(19\alpha-6\beta)^2$

First,

$19\alpha=19\cdot \frac{117}{19}=117$

and

$6\beta=6\cdot \frac{41}{2}=123.$

Thus,

$19\alpha-6\beta=117-123=-6.$

Therefore,

$(19\alpha-6\beta)^2=(-6)^2=36.$

6. Check with options

$36$ corresponds to Option C.

So the correct answer is C.