Square $ BDEC$ with center $ F$ is constructed to the out of triangle $ ABC$ such that $ \angle A \equal{} 90{}^\circ$, $ \left|AB\right| \equal{} \sqrt {12}$, $ \left|AC\right| \equal{} 2$. If $ \left[AF\right]\bigcap \left[BC\right] \equal{} \left\{G\right\}$ , then $ \left|BG\right|$ will be

$\textbf{(A)}\  6 \minus{} 2\sqrt {3} \qquad\textbf{(B)}\  2\sqrt {3} \minus{} 1 \qquad\textbf{(C)}\  2 \plus{} \sqrt {3} \\ \qquad\textbf{(D)}\  4 \minus{} \sqrt {3} \qquad\textbf{(E)}\  5 \minus{} 2\sqrt {2}$
1. **Identify the given information and set up the coordinate system:**
   - Given: $\angle A = 90^\circ$, $|AB| = \sqrt{12}$, $|AC| = 2$.
   - Place $A$ at the origin $(0,0)$, $B$ at $(0, \sqrt{12})$, and $C$ at $(2, 0)$.

2. **Construct the square $BDEC$ with center $F$:**
   - Since $BDEC$ is a square, $D$ and $E$ are the vertices such that $BD = DE = EC = BC$.
   - The center $F$ of the square is the midpoint of $BC$.

3. **Calculate the coordinates of $F$:**
   - The coordinates of $B$ are $(0, \sqrt{12})$ and $C$ are $(2, 0)$.
   - The midpoint $F$ of $BC$ is:
     \[
     F = \left( \frac{0 + 2}{2}, \frac{\sqrt{12} + 0}{2} \right) = (1, \sqrt{3})
     \]

4. **Determine the intersection point $G$ of $AF$ and $BC$:**
   - The line $AF$ passes through $(0,0)$ and $(1, \sqrt{3})$.
   - The slope of $AF$ is $\frac{\sqrt{3} - 0}{1 - 0} = \sqrt{3}$.
   - The equation of line $AF$ is $y = \sqrt{3}x$.
   - The line $BC$ is vertical at $x = 2$.

5. **Find the intersection point $G$:**
   - Substitute $x = 2$ into the equation of $AF$:
     \[
     y = \sqrt{3} \cdot 2 = 2\sqrt{3}
     \]
   - Thus, $G$ has coordinates $(2, 2\sqrt{3})$.

6. **Calculate the length $BG$:**
   - The coordinates of $B$ are $(0, \sqrt{12})$ and $G$ are $(2, 2\sqrt{3})$.
   - Use the distance formula:
     \[
     BG = \sqrt{(2 - 0)^2 + (2\sqrt{3} - \sqrt{12})^2}
     \]
   - Simplify the expression:
     \[
     BG = \sqrt{4 + (2\sqrt{3} - 2\sqrt{3})^2} = \sqrt{4 + 0} = 2
     \]

7. **Verify the calculation:**
   - The previous solution used a different approach, but the correct calculation shows $BG = 2$.

The final answer is $\boxed{2}$