Let A denote the matrix . The eigenvalues of A are the roots of the equation . If we try putting , then that equation becomes . The vector (x,y,z) will be an eigenvector for if . But that system of equations reduces to , which has two linearly independent solutions. Therefore is indeed an eigenvalue of A, with multiplicity at least 2. But the sum of the three eigenvalues of A is equal to the trace of A, which is . Therefore the third eigenvalue must be . Finally, the determinant of A is the product of its eigenvalues, namely .
I don't know if that counts as a proof "using ONLY properties of Det.", but at least it didn't involve expanding the determinant.