The position vectors of the points A and B with respect to O are 2 i ^ + 2 j ^ + k ^ and 2 i ^ + 4 j ^ + 4 k…

$\text{Here }\mathrm{OA}=\sqrt{2^2+2^2+1^2}=3, \mathrm{OB}=\sqrt{2^2+4^2+4^2}=6$

So, $\text{Internal bisector }OD\text{ divides }AB\text{ in the ratio} =1:2$, using angle bisector theorem

Using section formula, we get

$\text{Position vectors of }D\text{ are} (2,\frac{ 8}{3},2)$

$\therefore \left|OD\right|= \sqrt{ 4+\frac{ 64 }{9}+4 }=\frac{\sqrt{136}}{3}$