1. Use the angle bisector theorem in 3D triangle geometry

Since $D$ is the point where the internal bisector of $\angle BAC$ meets side $BC$, we have

$\frac{BD}{DC}=\frac{AB}{AC}.$

So first compute $AB$ and $AC$.

2. Find the side lengths $AB$ and $AC$

Given:

$A(1,3,2),\quad B(-2,8,0),\quad C(3,6,7).$

$\overrightarrow{AB}=(-2-1,\,8-3,\,0-2)=(-3,5,-2)$

$AB=\sqrt{(-3)^2+5^2+(-2)^2}=\sqrt{9+25+4}=\sqrt{38}$

$\overrightarrow{AC}=(3-1,\,6-3,\,7-2)=(2,3,5)$

$AC=\sqrt{2^2+3^2+5^2}=\sqrt{4+9+25}=\sqrt{38}$

Thus,

$AB=AC.$

Hence by angle bisector theorem,

$\frac{BD}{DC}=1,$

so $D$ is the midpoint of $BC$.

3. Find coordinates of midpoint $D$ of $BC$

$B(-2,8,0),\quad C(3,6,7)$

Therefore,

$D=\left(\frac{-2+3}{2},\frac{8+6}{2},\frac{0+7}{2}\right)=\left(\frac12,7,\frac72\right).$

4. Find vector $\overrightarrow{AD}$

$\overrightarrow{AD}=\left(\frac12-1,\,7-3,\,\frac72-2\right)=\left(-\frac12,4,\frac32\right).$

Also,

$\overrightarrow{AC}=(2,3,5).$

5. Projection length of $\overrightarrow{AD}$ on $\overrightarrow{AC}$

The scalar projection of $\overrightarrow{AD}$ on $\overrightarrow{AC}$ is

$\frac{\overrightarrow{AD}\cdot\overrightarrow{AC}}{|\overrightarrow{AC}|}.$

Compute the dot product:

$\overrightarrow{AD}\cdot\overrightarrow{AC}=\left(-\frac12\right)(2)+4(3)+\left(\frac32\right)(5)$

$=-1+12+\frac{15}{2}=11+\frac{15}{2}=\frac{37}{2}.$

And

$|\overrightarrow{AC}|=\sqrt{38}.$

So the required projection length is

$\frac{37/2}{\sqrt{38}}=\frac{37}{2\sqrt{38}}.$

6. Match with options

$\boxed{\frac{37}{2\sqrt{38}}}$

So the correct option is A.