Let O be the origin and the position vector of A and B be 2 ^ i + 2...

We know that, the internal bisector divides base in the same ratio as the two remaining sides.

$⇒ \dfrac{OA}{OB} = \dfrac{AC}{BC}$

$⇒ \dfrac{\sqrt{{2}^{2} + {2}^{2} + {1}^{2}}}{\sqrt{{2}^{2} + {4}^{2} + {4}^{2}}} = \dfrac{AC}{BC}$

$⇒ \dfrac{3}{6} = \dfrac{AC}{BC}$

$⇒ AC: BC = 1: 2$

$⇒ C ≡ \left(\dfrac{1 \times 2 + 2 \times 2}{3}, \dfrac{1 \times 4 + 2 \times 2}{3}, \dfrac{1 \times 4 + 2 \times 1}{3}\right)$

$⇒ C ≡ \left(2, \dfrac{8}{3}, 2\right)$

$⇒ OC = \sqrt{{2}^{2} + {\left(\dfrac{8}{3}\right)}^{2} + {2}^{2}}$

$⇒ OC = \sqrt{8 + \dfrac{64}{9}} = \dfrac{\sqrt{136}}{3}$

$⇒ OC = \dfrac{2\sqrt{34}}{3}$