A different (and quite easy) strategic approach to solving this problem without getting into square roots, the distance formula, or the Pythagorean theorem is to create a rectangle and subtract the areas of the right triangles not contained in the triangle you're looking for.
Specifically, make points at (-3, 11), (3,11) and (3,0), creating a rectangle with vertices at these three points and (-3,0).
The area of this rectangle is 6 x 11 = 66
Now subtract the area of the right triangle formed by (-3, 0), (-3,11) and (0,11), the area of which is (1/2)(11)(3) = 33/2
Then subtract the area of the right triangle formed by (0,11), (3,11) and (3,2), the area of which is (1/2)(9)(3) = 27/2
Then subtract the area of the right triangle formed by (-3, 0), (3,0), and (3,2), the area of which is (1/2)(6)(2) = 6
Your final answer will be 66 - (33/2 + 27/2 + 6), which is 66 - 36, which equals 30.
Easy, right? Draw it and you'll see.