We are given that $\sin \alpha - \cos \alpha = \frac{7}{13}$. We need to find the value of $A$, where $A = \frac{1 - 13|\sin \alpha + \cos \alpha|}{\sqrt{4 + 169\sin \alpha \cos \alpha}}$.

1. Problem Description

We are given that $\sin \alpha - \cos \alpha = \frac{7}{13}$. We need to find the value of $A$, where $A = \frac{1 - 13|\sin \alpha + \cos \alpha|}{\sqrt{4 + 169\sin \alpha \cos \alpha}}$.

2. Solution Steps

First, square the given equation:

$(\sin \alpha - \cos \alpha)^2 = (\frac{7}{13})^2$

$\sin^2 \alpha - 2\sin \alpha \cos \alpha + \cos^2 \alpha = \frac{49}{169}$

Since $\sin^2 \alpha + \cos^2 \alpha = 1$, we have:

$1 - 2\sin \alpha \cos \alpha = \frac{49}{169}$

$2\sin \alpha \cos \alpha = 1 - \frac{49}{169} = \frac{169 - 49}{169} = \frac{120}{169}$

$\sin \alpha \cos \alpha = \frac{60}{169}$

Next, consider $(\sin \alpha + \cos \alpha)^2$:

$(\sin \alpha + \cos \alpha)^2 = \sin^2 \alpha + 2\sin \alpha \cos \alpha + \cos^2 \alpha = 1 + 2\sin \alpha \cos \alpha$

$(\sin \alpha + \cos \alpha)^2 = 1 + 2(\frac{60}{169}) = 1 + \frac{120}{169} = \frac{169 + 120}{169} = \frac{289}{169}$

Therefore, $\sin \alpha + \cos \alpha = \pm\sqrt{\frac{289}{169}} = \pm\frac{17}{13}$.

Now we evaluate the denominator of A:

$\sqrt{4 + 169\sin \alpha \cos \alpha} = \sqrt{4 + 169(\frac{60}{169})} = \sqrt{4 + 60} = \sqrt{64} = 8$

We have two possible values for $\sin \alpha + \cos \alpha$, namely $\frac{17}{13}$ and $-\frac{17}{13}$.

Case 1: $\sin \alpha + \cos \alpha = \frac{17}{13}$

$A = \frac{1 - 13|\frac{17}{13}|}{8} = \frac{1 - 13(\frac{17}{13})}{8} = \frac{1 - 17}{8} = \frac{-16}{8} = -2$

Case 2: $\sin \alpha + \cos \alpha = -\frac{17}{13}$

$A = \frac{1 - 13|-\frac{17}{13}|}{8} = \frac{1 - 13(\frac{17}{13})}{8} = \frac{1 - 17}{8} = \frac{-16}{8} = -2$

In either case, $A = -2$.

3. Final Answer

-2