If $$x, y, z$$ are in arithmetic progression with common difference $$d$$, $$x \neq 3d$$, and the determinant of the matrix $$\begin{bmatrix} 3 & 4\sqrt{2} & x \\ 4 & 5\sqrt{2} & y \\ 5 & k & z \end{bmatrix}$$ is zero, then the value of $$k^2$$ is:

We are given that $$x, y, z$$ are in arithmetic progression with common difference $$d$$, so $$y = x + d$$ and $$z = x + 2d$$.

The determinant of the matrix $$\begin{vmatrix} 3 & 4\sqrt{2} & x \\ 4 & 5\sqrt{2} & y \\ 5 & k & z \end{vmatrix} = 0$$.

Since $$y = x + d$$ and $$z = x + 2d$$, we apply row operations $$R_2 \to R_2 - R_1$$ and $$R_3 \to R_3 - R_1$$:

$$\begin{vmatrix} 3 & 4\sqrt{2} & x \\ 1 & \sqrt{2} & d \\ 2 & k - 4\sqrt{2} & 2d \end{vmatrix} = 0$$.

Now apply $$R_3 \to R_3 - 2R_2$$:

$$\begin{vmatrix} 3 & 4\sqrt{2} & x \\ 1 & \sqrt{2} & d \\ 0 & k - 6\sqrt{2} & 0 \end{vmatrix} = 0$$.

Expanding along $$R_3$$: the only non-zero entry in the third row is in the second column. The cofactor expansion gives $$-(k - 6\sqrt{2}) \cdot \begin{vmatrix} 3 & x \\ 1 & d \end{vmatrix} = 0$$.

Computing the $$2 \times 2$$ determinant: $$\begin{vmatrix} 3 & x \\ 1 & d \end{vmatrix} = 3d - x$$.

So $$-(k - 6\sqrt{2})(3d - x) = 0$$.

We are given $$x \neq 3d$$, so $$3d - x \neq 0$$. Therefore $$k - 6\sqrt{2} = 0$$, giving $$k = 6\sqrt{2}$$.

Thus $$k^2 = (6\sqrt{2})^2 = 36 \times 2 = 72$$.

The answer is $$72$$, which is Option A.