Question goes as
If $\vec A$ and $\vec B$ are invariant under rotation, the prove that $ \vec A \times \vec B $ is also invariant.
However solution of on the other page is not given. Says that if you replace A and B with A' and B' and i,j,k with i', j', k' you will get the desired result.
I tried to change $ (A_2 B_3 - A_3 B_2 ) \hat i$ into $ A' $ and $ B'$ but I am getting zero.I would like to have a hint on this.
I'm not exactly sure what you are asking, however the following may be useful as a less verbose way of obtaining the same result.
The key fact is that the cross product of $A,B$ is the unique element $A\times B$ such that $\langle x, A\times B \rangle = \det \begin{bmatrix} A & B & x\end{bmatrix}$, $\forall x$.
Let $Q$ be a rotation (ie, $Q^TQ = I$), then using the properties of $\det$ we have $$\det \begin{bmatrix} A & B & x\end{bmatrix} =\det Q^T Q \det \begin{bmatrix} A & B & x\end{bmatrix} = \det Q \det \begin{bmatrix} QA & QB & Qx\end{bmatrix},$$ from which we obtain $\langle x, A\times B \rangle = \det Q \langle Qx, QA\times QB \rangle = \langle x, (\det Q) Q^T(QA\times QB) \rangle$. Since this is true for all $x$, we have $A\times B = (\det Q) Q^T(QA\times QB)$, or $$Q(A\times B) = (\det Q) QA\times QB.$$
Thus, if $Q$ is a proper rotation ($\det Q = +1$), you have $Q(A\times B) = QA\times QB$ (ie, the cross product is invariant under proper rotations).
Your question posits that both $A,B$ are invariant under rotation (which seems like a fairly restrictive condition), in which case $A = QA$, $B= QB$, from which it follows that $Q(A\times B) = A\times B$, ie, $A\times B$ is invariant under rotation too (assuming a proper rotation, of course).