$x^2+y^2+z^2=32$

for max z, x =y now 2x+z = 8 i.e. x = (8-z)/2 now 2x^2+z^2=32, solving z = 16/3

Unfortunately I don't know how to solve this algebraically (If there anyone who can give the algebraic solution to this problem please do so). So used my intuition instead. I thought that the answer for z should be between the two largest numbers in the choices given, namely: $\frac{16}{3}$ and $\frac{17}{3}$ . I know that the answer should be that when you square it is smaller than 32. Now checking the two choices, ( $\frac{17}{3}$ )^2=289/9=32.2222...... which is greater than 32 which means that this isn't the right answer. Checking $\frac{16}{3}$ , ( $\frac{16}{3}$ )^2= $\frac{256}{9}$ =28.444...... which satisfies the condition that the value of $\boxed{z^{2} <32}$ . Therefor the maximum value of $\boxed{z=16/3}$

it is better to be lucky than to be intelligent