For each of the three elementary row operations, applying it either does nothing to the determinant (in the case of a $R_1 \mapsto R_1 + 6 R_3$ sort of operation) or it multiplies by a non-zero scalar ($-1$ for a flip-flop, and whatever scalar you're multiplying the row by for a scalar multiplication).
So: elementary row operations can change the determinant of a matrix, but they can't change whether that determinant is zero or not.
Therefore, given a square matrix with $\det A = 0$, when we echellonize it, we must get a matrix with some $0$s on the diagonal (because a diagonal matrix with no $0$s on the diagonal doesn't have determinant $0$). It's then obvious that the corresponding homogeneous equation has a non-zero solution (corresponding to the row of $0$s in the matrix).
Conversely, if there exists a non-zero solution, then our matrix echellonizes to a matrix with a row full of zeroes, so the determinant of that matrix is zero; hence the determinant of our original matrix was zero.