$ a$ is a real number. $ x_1$ and $ x_2$ are the distinct roots of $ x^2 \plus{} ax \plus{} 2 \equal{} x$. $ x_3$ and $ x_4$ are the distinct roots of $ (x \minus{} a)^2 \plus{} a(x \minus{} a) \plus{} 2 \equal{} x$. If $ x_3 \minus{} x_1 \equal{} 3(x_4 \minus{} x_2)$, then $ x_4 \minus{} x_2$ will be ?

$\textbf{(A)}\  \frac {a}{2} \qquad\textbf{(B)}\  \frac {a}{3} \qquad\textbf{(C)}\  \frac {2a}{3} \qquad\textbf{(D)}\  \frac {3a}{2} \qquad\textbf{(E)}\ \text{None}$
1. We start with the given equations:
   \[
   x^2 + ax + 2 = x
   \]
   and
   \[
   (x - a)^2 + a(x - a) + 2 = x.
   \]

2. Simplify the first equation:
   \[
   x^2 + ax + 2 = x \implies x^2 + (a-1)x + 2 = 0.
   \]
   Let the roots of this equation be \(x_1\) and \(x_2\). By Vieta's formulas, we have:
   \[
   x_1 + x_2 = -(a-1) = 1 - a
   \]
   and
   \[
   x_1 x_2 = 2.
   \]

3. Simplify the second equation:
   \[
   (x - a)^2 + a(x - a) + 2 = x \implies x^2 - 2ax + a^2 + ax - a^2 + 2 = x \implies x^2 - (a+1)x + 2 = 0.
   \]
   Let the roots of this equation be \(x_3\) and \(x_4\). By Vieta's formulas, we have:
   \[
   x_3 + x_4 = a + 1
   \]
   and
   \[
   x_3 x_4 = 2.
   \]

4. We are given the condition:
   \[
   x_3 - x_1 = 3(x_4 - x_2).
   \]

5. Express \(x_1\) and \(x_3\) in terms of \(x_2\) and \(x_4\):
   \[
   x_1 = 1 - a - x_2
   \]
   and
   \[
   x_3 = a + 1 - x_4.
   \]

6. Substitute these into the given condition:
   \[
   (a + 1 - x_4) - (1 - a - x_2) = 3(x_4 - x_2).
   \]

7. Simplify the equation:
   \[
   a + 1 - x_4 - 1 + a + x_2 = 3(x_4 - x_2) \implies 2a - x_4 + x_2 = 3(x_4 - x_2).
   \]

8. Combine like terms:
   \[
   2a - x_4 + x_2 = 3x_4 - 3x_2 \implies 2a = 4x_4 - 4x_2 \implies 4(x_4 - x_2) = 2a \implies x_4 - x_2 = \frac{a}{2}.
   \]

Conclusion:
\[
\boxed{\frac{a}{2}}
\]