Given,

$A\left(1,3,2\right), B\left(-2,8,0\right) \& C\left(3,6,7\right)$

And angle bisector of $\angle BAC$ meets the line $BC$ at $D$,

Now, plotting the diagram we get,

Now, finding

$\overrightarrow{AC}=2\hat{i}+3\hat{j}+5\hat{k}$

$\Rightarrow \left|\overrightarrow{AC}\right|=\sqrt{4+9+25}=\sqrt{38}$

And $\left|\overrightarrow{AB}\right|=\sqrt{9+25+4}=\sqrt{38}$

Using angle bisector theorem, we get

$\frac{BD}{CD}=\frac{AB}{AC}$

$\Rightarrow BD:CD=1:1$

Using section formula to find coordinates of $D$,

$\Rightarrow D\equiv \left(\frac{1\times 3+1\left(-2\right)}{1+2}, \frac{1\times 6+1\times 8}{1+2}, \frac{1\times 7+1\times 0}{1+2}\right)$

$\Rightarrow D\equiv \left(\frac{1}{2}, 7, \frac{7}{2}\right)$

$\Rightarrow \overrightarrow{AD}=-\frac{1}{2}\hat{i}+4\hat{j}+\frac{3}{2}\hat{k}$

$\Rightarrow \overrightarrow{AD}=\frac{1}{2}\left(-\hat{i}+8\hat{j}+3\hat{k}\right)$

Length of projection of $\overrightarrow{AD}$ on $\overrightarrow{AC}$ is given by,

$\Rightarrow d=\left|\frac{\overrightarrow{AD}·\overrightarrow{AC}}{\overrightarrow{AC}}\right|$

$\Rightarrow d=\frac{37}{2\sqrt{38}}$