The position vectors of the vertices A , B and C of a triangle are 2 i ^ - 3 j ^ + 3 k ^ , 2 i ^ + 2 j ^ + 3…

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

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

$\Rightarrow \mathrm{AB}=5$

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

$\Rightarrow \mathrm{AC}=\sqrt{9+16}$

$\Rightarrow \mathrm{AC}=5$

Also, $\mathrm{AD}$ is the angle bisector of $\angle \mathrm{BAC}$.

$\Rightarrow \frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{BD}}{\mathrm{CD}}$

$\Rightarrow \mathrm{BD}=\mathrm{CD}$

$\therefore$ $\mathrm{D}$ is midpoint of $\mathrm{BC}$

So, using mid-point formula,

$\Rightarrow \mathrm{D}\equiv \left(\frac{2-1}{2},\frac{2+1}{2},\frac{3+3}{2}\right)$

$\Rightarrow \mathrm{D}\equiv \left(\frac{1}{2},\frac{3}{2},3\right)$

$\Rightarrow \mathrm{AD}=l=\sqrt{{\left(2-\frac{1}{2}\right)}^2+{\left(-3-\frac{3}{2}\right)}^2+(3-3{)}^2}$

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

$\Rightarrow l=\sqrt{\frac{9}{4}+\frac{81}{4}}$

$\Rightarrow l=\sqrt{\frac{45}{2}}$

$\Rightarrow l^2=\frac{45}{2}$

$\Rightarrow 2l^2=45$