Let $A$ and $B$ two bounded operators on a complex Hilbert space $E$. We define \begin{eqnarray*} \Omega(A,B)=\sup\left \{ \| \alpha A+\beta B\| \,;\;\; \alpha,\beta\in \mathbb{C},\; |\alpha|^2+|\beta|^2\leq 1 \right\}. \end{eqnarray*} We can prove that $\Omega(A,B)$ defines a norm on the bounded operators on $E\times E$.
I want to calculate $\Omega(A,B)$ when $$A=\begin{pmatrix}0&0&1\\0&0&0\\0&0&0\end{pmatrix}\quad\text{ and }\quad B=\begin{pmatrix}0&1&0\\0&0&0\\0&0&0\end{pmatrix}.$$
We can check that \begin{align*} \Omega(A,B)&= \sup_{|\alpha|^2+|\beta|^2\leq 1}\sup_{|x|^2+|y|^2+|z|^2\leq 1} \Big\|\begin{pmatrix} 0&\beta&\alpha\\0&0&0\\0&0&0 \end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}\Big\| \\ &=\sup_{|\alpha|^2+|\beta|^2\leq 1}\sup_{|x|^2+|y|^2+|z|^2\leq 1} |\alpha z+\beta y| \end{align*}
I think we can proceed as follows \begin{align*} |\alpha z+\beta y| &\leq \sqrt{\left(|\alpha| |z|+|\beta| |y|\right)^2}\\ &\leq \sqrt{\left(|\alpha|^2 +|\beta|^2 \right)\left(|y|^2 +|z|^2 \right)}\\ &\leq \sqrt{\left(|\alpha|^2 +|\beta|^2 \right)\left(|x|^2 +|y|^2+ |z|^2\right)}\\ &\leq 1 \end{align*} So, $$\Omega(A,B)\leq1.$$
Is $\Omega(A,B)=1$?
Computing a lower bound for norms which involve a supremum is usually done by evaluating the supremum at a "suitable" point. In this case one possible way is $$ \Omega(A,B)=\sup_{|\alpha|^2+|\beta|^2\leq1}\sup_{|x|^2+|y|^2+|z|^2\leq1}|\alpha z+\beta y|\overset{(x,y,z)=(0,0,1)}\geq\sup_{|\alpha|^2+|\beta|^2\leq1}|\alpha|\overset{(\alpha,\beta)=(1,0)}\geq 1 $$ which -- together with the inequality $\Omega(A,B)\leq 1$ which you already proved -- shows $\Omega(A,B)=1$ as desired.