This is a very simple exercise of the Pythagorean Theorem. The tricky part is the FOIL.

\(a^2 + b^2 = c^2\)

\((3x+3)^2 + (x+2)^2 = (2x+7)^2\)

\((9x^2 + 18x + 9) + (x^2 + 4x + 4) = (4x^2 + 28x + 49)\)

\(10x^2 + 22x + 13 = 4x^2 + 28x + 49.\)

\(6x^2 -6x - 36 = 0\)

\(6(x^2 - x - 6) = 0\)

\(6(x-3)(x+2) = 0\)

\(x = 3\) or \(x = -2.\)

Since x can't be negative, we know that x=3.
Answer: B

OR, the ADVANCED GMATer would notice that all of the answers are integers that are easily manageable. A quicker approach would be to plug these values in and find a known Pythagorean triple.

x = 2 results in (4,9,11). Not a Pythagorean triple.
x = 3 results in (5,12,13). This is a known Pythagorean triple and we can stop here. We have found our answer:

Answer B.