The position vectors of the vertices A, B and C of a triangle are 2...

Given: $A\left(2, - 3, 3\right), B\left(2, 2, 3\right)\text{and}C\left(- 1, 1, 3\right)$ are vertices of a $△ ABC$.

$AB = \sqrt{{\left(2 - 2\right)}^{2} + {\left(2 + 3\right)}^{2} + {\left(3 - 3\right)}^{2}}$

$⇒ AB = 5$

$⇒ AC = \sqrt{{\left(2 + 1\right)}^{2} + {\left(- 3 - 1\right)}^{2} + {\left(3 - 3\right)}^{2}}$

$⇒ AC = \sqrt{9 + 16}$

$⇒ AC = 5$

Also, $AD$ is the angle bisector of $∠ BAC$.

$⇒ \dfrac{AB}{AC} = \dfrac{BD}{CD}$

$⇒ BD = CD$

$∴$ $D$ is midpoint of $BC$

So, using mid-point formula,

$⇒ D ≡ \left(\dfrac{2 - 1}{2}, \dfrac{2 + 1}{2}, \dfrac{3 + 3}{2}\right)$

$⇒ D ≡ \left(\dfrac{1}{2}, \dfrac{3}{2}, 3\right)$

$⇒ AD = l = \sqrt{{\left(2 - \dfrac{1}{2}\right)}^{2} + {\left(- 3 - \dfrac{3}{2}\right)}^{2} + ( 3 - 3{)}^{2}}$

$⇒ l = \sqrt{{\left(\dfrac{3}{2}\right)}^{2} + {\left(- \dfrac{9}{2}\right)}^{2}}$

$⇒ l = \sqrt{\dfrac{9}{4} + \dfrac{81}{4}}$

$⇒ l = \sqrt{\dfrac{45}{2}}$

$⇒ {l}^{2} = \dfrac{45}{2}$

$⇒ 2{l}^{2} = 45$