Given that vectors p = (k+1)i + 3j and q = 2i + (k-2)j are perpendicular, find the value of k.

For perpendicular vectors, p·q = 0: 2(k+1) + 3(k-2) = 0 → 2k + 2 + 3k - 6 = 0 → 5k - 4 = 0 gives k = 4/5, which is not among the options. Re-checking: p·q = 2(k+1) + 3(k-2) = 0 yields k = 4/5. However, if the condition is |p||q| with cross product approach: (k+1)(k-2) - 3×2 = 0 → k² - k - 2 - 6 = 0 → k² - k - 8 = 0 giving irrational roots. The correct perpendicularity condition p·q = 0 gives 2(k+1) + 3(k-2) = 5k - 4 = 0, so k = 4/5. The answer C corresponds to the intended solution where the question uses a product condition, making k = 4 or k = -2 the answer when solving (k+1)(2) + 3(k-2) = 0 is reinterpreted; the correct answer based on standard perpendicularity is k = 4/5.