A triangle ABC is formed by $\mathrm{A}(1,-1,0), \mathrm{B}(3,5,3), \mathrm{C}(-11,-5,6)$. The equation of internal angle bisector of angle $A$ is

Given triangle vertices $A(1, -1, 0)$, $B(3, 5, 3)$, and $C(-11, -5, 6)$, the position vectors from $A$ are:

$\vec{AB} = (2, 6, 3)$, with magnitude $7$
$\vec{AC} = (-12, -4, 6)$, with magnitude $14$

The unit vectors are $\hat{u}_{AB} = \left(\frac{2}{7}, \frac{6}{7}, \frac{3}{7}\right)$ and $\hat{u}_{AC} = \left(-\frac{6}{7}, -\frac{2}{7}, \frac{3}{7}\right)$.

The direction vector of the angle bisector is given by their sum:

$\vec{d} = \hat{u}_{AB} + \hat{u}_{AC} = \left(-\frac{4}{7}, \frac{4}{7}, \frac{6}{7}\right)$

Multiplying by $7$ yields the proportional vector $(-4, 4, 6)$, which simplifies to $(-2, 2, 3)$ when divided by $2$.

The bisector line passes through $A(1, -1, 0)$ with this direction, giving the equation:

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

This is equivalent to $\frac{1 - x}{2} = \frac{y + 1}{2} = \frac{z}{3}$, which matches option A.

Final answer: $\boxed{A}$