The position vector of vertices A, B, C of Δ are $\stackrel{^<!--\; ^\; -->}{i}+2\stackrel{^<!--\; ^\; -->}{j}+3\stackrel{^<!--\; ^\; -->}{k},\stackrel{^<!--\; ^\; -->}{i}+\stackrel{^<!--\; ^\; -->}{j}+3\stackrel{^<!--\; ^\; -->}{k},2\stackrel{^<!--\; ^\; -->}{i}+\stackrel{^<!--\; ^\; -->}{j}+3\stackrel{^<!--\; ^\; -->}{k}$ respectively. Let l is the length of angle bisector of ∠BAC , then the value of l² is

The correct option is 4 - 2√2.

From the given information and the provided diagram, we are given a triangle $ABC$ where the position vectors of the vertices are:
Vertex $A=\hat{i}+2\hat{j}+3\hat{k}$, which corresponds to the coordinates $A(1,2,3)$.
Vertex $B=\hat{i}+\hat{j}+3\hat{k}$, which corresponds to the coordinates $B(1,1,3)$.
Vertex $C=2\hat{i}+\hat{j}+3\hat{k}$, which corresponds to the coordinates $C(2,1,3)$.

Step 1: Find the lengths of the sides AB and AC
The length of side $AB$ is calculated using the distance formula:
$AB=\sqrt{{(1-1)}^2+{(1-2)}^2+{(3-3)}^2}=\sqrt{0+1+0}=1$
Similarly, the length of side $AC$ is:
$AC=\sqrt{{(2-1)}^2+{(1-2)}^2+{(3-3)}^2}=\sqrt{1+1+0}=\sqrt{2}$

Step 2: Apply the Angle Bisector Theorem
Let the angle bisector of $\angle BAC$ meet the side $BC$ at a point $D$.
According to the Angle Bisector Theorem, the bisector of an angle divides the opposite side in the ratio of the lengths of the other two sides. Therefore:
$\frac{BD}{DC}=\frac{AB}{AC}=\frac{1}{\sqrt{2}}$
This shows that point $D$ divides the line segment joining $B(1,1,3)$ and $C(2,1,3)$ internally in the ratio $1:\sqrt{2}$.

Step 3: Determine the coordinates of point D
Using the section formula, the coordinates of $D$ are:
$D=(\frac{1·2+\sqrt{2}·1}{1+\sqrt{2}},\frac{1·1+\sqrt{2}·1}{1+\sqrt{2}},\frac{1·3+\sqrt{2}·3}{1+\sqrt{2}})$
Simplifying each coordinate:
$D=(\frac{2+\sqrt{2}}{\sqrt{2}+1},1,3)$

Step 4: Find the length of the angle bisector AD (l)
Let $l$ be the length of $AD$. The distance between $A(1,2,3)$ and $D(\frac{2+\sqrt{2}}{\sqrt{2}+1},1,3)$ is given by:
$l=\sqrt{{(\frac{2+\sqrt{2}}{\sqrt{2}+1}-1)}^2+{(1-2)}^2+{(3-3)}^2}$
Let us simplify the term $\frac{2+\sqrt{2}}{\sqrt{2}+1}-1$:
$\frac{2+\sqrt{2}}{\sqrt{2}+1}-1=\frac{2+\sqrt{2}-(\sqrt{2}+1)}{\sqrt{2}+1}=\frac{1}{\sqrt{2}+1}$
Substituting this back into the expression for $l$:
$l=\sqrt{{\left(\frac{1}{\sqrt{2}+1}\right)}^2+{(-1)}^2+0}$
Squaring both sides to find $l^2$:
$l^2={\left(\frac{1}{\sqrt{2}+1}\right)}^2+1$

Step 5: Simplify the final expression
Rationalize the denominator of the fraction:
$\frac{1}{\sqrt{2}+1}=\frac{\sqrt{2}-1}{(\sqrt{2}+1)(\sqrt{2}-1)}=\frac{\sqrt{2}-1}{2-1}=\sqrt{2}-1$
Now, substitute this value into the equation for $l^2$:
$l^2={(\sqrt{2}-1)}^2+1$
Expand the squared term:
$l^2=(2-2\sqrt{2}+1)+1$
$l^2=4-2\sqrt{2}$