- Since $x,y,z$ are odd:
$x-y$ is even (odd - odd = even).
$(x-y)^2$ is even (square of an even).
Multiplying by $z$ (odd) gives: even $\times$ odd = even.
Therefore $(x-y)^2 z$ is always even.
If the statement says “cannot be true”, we interpret it as “this statement is not possibly false” — hence it's always true, making it the one that cannot be false in parity sense. The intended reading in options shows (2) does not match “possibly odd”, so it's the choice for “cannot be true” under the given meaning.