The total number of matrices $$A = \begin{pmatrix} 0 & 2y & 1 \\ 2x & y & -1 \\ 2x & -y & 1 \end{pmatrix}$$, $$(x, y \in R, x \neq y)$$ for which $$A^TA = 3I_3$$ is:

We are given the matrix $$A = \begin{pmatrix} 0 & 2y & 1 \\ 2x & y & -1 \\ 2x & -y & 1 \end{pmatrix}$$ and the condition $$A^T A = 3I_3$$.

Step 1: Compute $$A^T A$$

The transpose of $$A$$ is:

$$A^T = \begin{pmatrix} 0 & 2x & 2x \\ 2y & y & -y \\ 1 & -1 & 1 \end{pmatrix}$$

The condition $$A^T A = 3I_3$$ means the columns of $$A$$ are orthogonal and each column has magnitude $$\sqrt{3}$$.

Step 2: Column norms equal $$\sqrt{3}$$

Column 1: $$(0, 2x, 2x)^T$$. Norm squared: $$0 + 4x^2 + 4x^2 = 8x^2 = 3$$

$$x^2 = \frac{3}{8}$$, so $$x = \pm\sqrt{\frac{3}{8}}$$

Column 2: $$(2y, y, -y)^T$$. Norm squared: $$4y^2 + y^2 + y^2 = 6y^2 = 3$$

$$y^2 = \frac{1}{2}$$, so $$y = \pm\frac{1}{\sqrt{2}}$$

Column 3: $$(1, -1, 1)^T$$. Norm squared: $$1 + 1 + 1 = 3$$ ✔

Step 3: Verify orthogonality

Column 1 $$\cdot$$ Column 2: $$0 \cdot 2y + 2x \cdot y + 2x \cdot (-y) = 0 + 2xy - 2xy = 0$$ ✔

Column 1 $$\cdot$$ Column 3: $$0 \cdot 1 + 2x \cdot (-1) + 2x \cdot 1 = -2x + 2x = 0$$ ✔

Column 2 $$\cdot$$ Column 3: $$2y \cdot 1 + y \cdot (-1) + (-y) \cdot 1 = 2y - y - y = 0$$ ✔

All orthogonality conditions are automatically satisfied, regardless of the signs of $$x$$ and $$y$$.

Step 4: Check the constraint $$x \neq y$$

We have $$x = \pm\sqrt{3/8}$$ and $$y = \pm 1/\sqrt{2}$$.

Since $$\sqrt{3/8} = \frac{\sqrt{3}}{2\sqrt{2}} \approx 0.612$$ and $$\frac{1}{\sqrt{2}} \approx 0.707$$, these values are never equal.

So all 4 combinations of signs satisfy $$x \neq y$$.

Step 5: Count the matrices

There are 2 choices for $$x$$ and 2 choices for $$y$$, giving $$2 \times 2 = 4$$ matrices.

The correct answer is Option C: 4.