Step 1: Write the potential as a matrix.
The potential part of $H$ is $V(x,y) = \tfrac12 m\Omega^2 x^2 + \tfrac12 m\Omega^2 y^2 + m\omega^2 xy$. This can be written as $V = \tfrac12 m \begin{pmatrix} x & y \end{pmatrix} M \begin{pmatrix} x \\ y \end{pmatrix}$ with the symmetric matrix $M = \begin{pmatrix} \Omega^2 & \omega^2 \\ \omega^2 & \Omega^2 \end{pmatrix}$, since expanding this product gives back $\Omega^2x^2 + \Omega^2y^2 + 2\omega^2 xy$, matching twice the cross term coefficient we need.
Step 2: Find the eigenvalues of $M$.
The normal mode frequencies squared are the eigenvalues of $M$. Solve $\det(M - \lambda I) = 0$:
\[ (\Omega^2-\lambda)^2 - \omega^4 = 0 \]
\[ \Omega^2 - \lambda = \pm\omega^2 \]
So $\lambda_1 = \Omega^2 + \omega^2$ and $\lambda_2 = \Omega^2 - \omega^2$. These are the squares of the two normal mode frequencies, $\omega_1 = \sqrt{\Omega^2+\omega^2}$ and $\omega_2 = \sqrt{\Omega^2-\omega^2}$. The condition $\omega < \Omega$ keeps $\lambda_2$ positive, so both frequencies stay real.
Step 3: Note that the kinetic term stays diagonal under this rotation.
The eigenvectors of a real symmetric $2\times2$ matrix with equal diagonal entries are always the $45^\circ$ rotated directions, exactly the $x_1, x_2$ combination given in the question. Since this eigenvector rotation is orthogonal, it leaves the kinetic energy $-\tfrac{\hbar^2}{2m}(\partial_x^2+\partial_y^2)$ unchanged in form, so in the rotated frame the full Hamiltonian splits into two separate one-dimensional oscillators with frequencies $\omega_1$ and $\omega_2$.
Step 4: Add the two zero-point energies.
Each one-dimensional harmonic oscillator contributes a ground state energy of $\tfrac12\hbar\omega_i$. Adding both normal modes:
\[ E_0 = \frac12\hbar\left(\sqrt{\Omega^2+\omega^2} + \sqrt{\Omega^2-\omega^2}\right) \]
Step 5: Rule out the other options.
Option (B) drops the $\tfrac12$ that comes from summing two zero-point energies of $\tfrac12\hbar\omega_i$ each. Options (C) and (D) would only appear if the eigenvalues of $M$ had come out as $\Omega^2 \pm \Omega\omega$, which is not what solving $\det(M-\lambda I)=0$ gives here.
Final Answer:
The eigenvalue approach gives the same result, $E_0 = \dfrac{\hbar}{2}\big[\sqrt{\Omega^2-\omega^2}+\sqrt{\Omega^2+\omega^2}\big]$, option (A).\[ \boxed{E_0 = \frac{\hbar}{2}\left[\sqrt{\Omega^2-\omega^2}+\sqrt{\Omega^2+\omega^2}\right]} \]