Write the given position vectors

$\vec{OA}=2\hat i+2\hat j+\hat k=(2,2,1),\qquad \vec{OB}=2\hat i+4\hat j+4\hat k=(2,4,4)$

Use the angle bisector theorem in triangle $AOB$

Since $OC$ is the internal bisector of $\angle AOB$ and meets side $AB$ at $C$, in triangle $AOB$ we have

$\frac{AC}{CB}=\frac{OA}{OB}$

First compute the lengths:

$OA=\sqrt{2^2+2^2+1^2}=\sqrt{9}=3$

$OB=\sqrt{2^2+4^2+4^2}=\sqrt{36}=6$

Hence,

$\frac{AC}{CB}=\frac{3}{6}=\frac{1}{2}$

So $C$ divides $AB$ internally in the ratio $1:2$.

Find the position vector of $C$ using section formula

If $C$ divides $AB$ internally in the ratio $1:2$, then

$\vec{OC}=\frac{1\cdot \vec{OB}+2\cdot \vec{OA}}{1+2}$

$\vec{OC}=\frac{(2,4,4)+2(2,2,1)}{3}$

$=\frac{(2,4,4)+(4,4,2)}{3}$

$=\frac{(6,8,6)}{3}=(2,\tfrac83,2)$

Compute the length $OC$

$OC=\sqrt{2^2+\left(\frac83\right)^2+2^2}$

$=\sqrt{4+\frac{64}{9}+4}$

$=\sqrt{8+\frac{64}{9}}$

$=\sqrt{\frac{72+64}{9}}=\sqrt{\frac{136}{9}}=\frac{\sqrt{136}}{3}$

$OC=\frac{2\sqrt{34}}{3}$

Match with the options

$\frac{2\sqrt{34}}{3}=\frac{2}{3}\sqrt{34}$

Therefore, the correct option is C.