Let $c=(-5,-4,5)$. Find the supporting hyperplane $H_c$ of the cross polytope $$P=\{(x_{1},x_{2},x_{3}) : |x_{1}|+|x_{2}|+|x_{3}| \le 1\}$$
I understand that we need to find $\delta=max\{
So far I have that $\delta=max\{-5x_{1}-4x_{2}+4x_{3} | x \in P\}$
You are looking for one of the 2 supporting planes with equations:
$$ux_1+vx_2+wx_3-d=0 \ \ \text{which is equivalent to} \ \ (u,v,w).(x_1,x_2,x_3)=d$$
in the direction $(-5,-4,5)$ which a normal direction, with respect to the unit octahedron (https://en.wikipedia.org/wiki/Octahedron) which is the classical name for this volume (I had never heard about the expression "cross polytope" for this volume). Thus the equation of the plane is
$$\tag{1}-5x_1-4x_2+5x_3=d$$
You can imagine (1) as the common equation of parallel planes sweeping through the octahedron, i.e., intersecting it for a certain range of values ($d_{min} \leq d \leq d_{max}$).
Now, what are these extremal solutions $d_{min}$ and $d_{max}$?
Inverting (1) in order to consider $d$ as a function of $x_1,x_2,x_3$,:
$$\tag{2}d=-5x_1-4x_2+5x_3$$
The fundamental remark is that extremal values of $d$ are taken on a vertex (classical result for a linear function with respect to a polyhedron.)
Thus it suffices to consider the six vertices:
$$\begin{array}{|rrr|} \hline1&0&0\\\hline0&1&0\\\hline0&0&1\\\hline-1&0&0\\\hline0&-1&0\\\hline0&0&-1\\\hline \end{array}$$
and evaluate expression (2) of $d$ on each of them, keeping the two extremal values of $d$, i.e., $d_{min}$ and $d_{max}$ that will be plugged into (1) for the final answer.
It remains for you to find $d_{min}$ and $d_{max}...$