Let $$O$$ be the origin and the position vector of $$A$$ and $$B$$ be $$2\hat{i} + 2\hat{j} + \hat{k}$$ and $$2\hat{i} + 4\hat{j} + 4\hat{k}$$ respectively. If the internal bisector of $$\angle AOB$$ meets the line $$AB$$ at $$C$$, then the length of $$OC$$ is

$$\vec{OA} = 2\hat{i} + 2\hat{j} + \hat{k}$$, so $$|\vec{OA}| = \sqrt{4+4+1} = 3$$

$$\vec{OB} = 2\hat{i} + 4\hat{j} + 4\hat{k}$$, so $$|\vec{OB}| = \sqrt{4+16+16} = 6$$

Apply the angle bisector theorem.

By the angle bisector theorem, the internal bisector of $$\angle AOB$$ divides $$AB$$ in the ratio $$OA:OB = 3:6 = 1:2$$.

Using the section formula (internal division in ratio $$1:2$$):

$$\vec{OC} = \frac{1 \cdot \vec{OB} + 2 \cdot \vec{OA}}{1 + 2} = \frac{(2\hat{i}+4\hat{j}+4\hat{k}) + 2(2\hat{i}+2\hat{j}+\hat{k})}{3}$$

$$= \frac{(2+4)\hat{i} + (4+4)\hat{j} + (4+2)\hat{k}}{3} = \frac{6\hat{i} + 8\hat{j} + 6\hat{k}}{3} = 2\hat{i} + \frac{8}{3}\hat{j} + 2\hat{k}$$

$$|OC| = \sqrt{4 + \frac{64}{9} + 4} = \sqrt{8 + \frac{64}{9}} = \sqrt{\frac{72 + 64}{9}} = \sqrt{\frac{136}{9}} = \frac{\sqrt{136}}{3} = \frac{2\sqrt{34}}{3}$$