If x ,   y ,   z are in arithmetic progression with common difference d ,   x ≠ 3 d ,…

Given $x, y, z$ are in arithmetic progression, hence $2y=x+z,$ also $\left|\begin{array}{ccc}3& 4\sqrt{2}& x\\ 4& 5\sqrt{2}& y\\ 5& k& z\end{array}\right|=0$

Applying $R_2\to R_1+R_3-2R_2$

$\Rightarrow \left|\begin{array}{ccc}3& 4\sqrt{2}& x\\ 0& k-6\sqrt{2}& 0\\ 5& k& z\end{array}\right|=0$

$\Rightarrow (k-6\sqrt{2})(3z-5x)=0$

$\Rightarrow (k-6\sqrt{2})=0$ or $(3z-5x)=0$

If $3z-5x=0$

$\Rightarrow 3(x+2d)-5x=0$

$\Rightarrow x=3d$ (Not possible)

$\Rightarrow k=6\sqrt{2}$

$\Rightarrow k^2=72.$