Question

Suppose the matrix A has eigenvalues lambda_1 = -1, lambda_2 = 1, lambda_3 = 2, with corresponding eigenvectors v_1 = [0 5 3]^T, v_2 = [2 0 1]^T, v_3 = [1 -1 0]^T. If you diagonalize A as A = PDP^-1 with P = [2 2 0; p_21 p_22 2; p_31 p_32 p_33], D = [2 0 0, 0 1 0; 0 0 -1], then p_32 = 1. Please explain how p_32 = 1 is found.