To solve the expression given in the image, we will approach it step-by-step:

\[ A = \left(\frac{15}{\sqrt{6}+1} + \frac{2}{\sqrt{6}-2} - \frac{6}{3-\sqrt{6}}\right) (2\sqrt{6} + 7) \]

### Step 1: Rationalize the denominators
For each fraction, we will multiply the numerator and the denominator by the conjugate of the denominator to rationalize it.

1. \(\frac{15}{\sqrt{6} + 1}\):

\[
\frac{15}{\sqrt{6} + 1} \cdot \frac{\sqrt{6} - 1}{\sqrt{6} - 1} = \frac{15(\sqrt{6} - 1)}{(\sqrt{6})^2 - 1^2} = \frac{15(\sqrt{6} - 1)}{6 - 1} = \frac{15(\sqrt{6} - 1)}{5} = 3\sqrt{6} - 3
\]

2. \(\frac{2}{\sqrt{6} - 2}\):

\[
\frac{2}{\sqrt{6} - 2} \cdot \frac{\sqrt{6} + 2}{\sqrt{6} + 2} = \frac{2(\sqrt{6} + 2)}{(\sqrt{6})^2 - 2^2} = \frac{2(\sqrt{6} + 2)}{6 - 4} = \frac{2(\sqrt{6} + 2)}{2} = \sqrt{6} + 2
\]

3. \(\frac{6}{3 - \sqrt{6}}\):

\[
\frac{6}{3 - \sqrt{6}} \cdot \frac{3 + \sqrt{6}}{3 + \sqrt{6}} = \frac{6(3 + \sqrt{6})}{3^2 - (\sqrt{6})^2} = \frac{6(3 + \sqrt{6})}{9 - 6} = \frac{6(3 + \sqrt{6})}{3} = 2(3 + \sqrt{6}) = 6 + 2\sqrt{6}
\]

### Step 2: Combine the fractions
Now substituting back into \(A\):

\[
A = (3\sqrt{6} - 3 + \sqrt{6} + 2 - 6 - 2\sqrt{6})(2\sqrt{6} + 7)
\]

### Step 3: Simplify the combined fraction.
Combine like terms:

\[
A = (3\sqrt{6} + \sqrt{6} - 2\sqrt{6} - 3 + 2 - 6) (2\sqrt{6} + 7)
\]
\[
A = (\sqrt{6} - 7) (2\sqrt{6} + 7)
\]

### Step 4: Distribute and Combine like terms.
Now distribute:

\[
A = \sqrt{6}(2\sqrt{6}) + \sqrt{6}(7) - 7(2\sqrt{6}) - 7(7)
\]

\[
A = 2(\sqrt{6})^2 + 7\sqrt{6} - 14\sqrt{6} - 49
\]

\[
A = 2 * 6 + 7\sqrt{6} - 14\sqrt{6} - 49
\]

\[
A = 12 + 7\sqrt{6} - 14\sqrt{6} - 49
\]

\[
A = 12 - 7\sqrt{6} - 49
\]

\[
A = -37 - 7\sqrt{6}
\]

Thus, the value of \(A\) is:

\[
A = -37 - 7\sqrt{6}
\]