Let $$A(x,y,Z)$$ be a point in $$xy-plain$$,which is equidistant from three points (0, 3, 2), (2, 0, 3) and (0, 0, 1). Let B = (1, 4, −1) and C = (2, 0, −2). Then among the statements $$(SI): \triangle ABC$$ is an isosceles right angled triangle, and (SI):the area of $$\triangle ABC$$ is $$\frac{9\sqrt{2}}{2}$$,

Given that point $$A(x, y, z)$$ lies in the xy-plane, $$z = 0$$, so $$A = (x, y, 0)$$. It is equidistant from points $$P(0, 3, 2)$$, $$Q(2, 0, 3)$$, and $$R(0, 0, 1)$$.

The distance formula states that the distance between two points $$(x_1, y_1, z_1)$$ and $$(x_2, y_2, z_2)$$ is $$\sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2 + (z_1 - z_2)^2}$$.

Setting $$AP = AR$$:

$$\sqrt{(x - 0)^2 + (y - 3)^2 + (0 - 2)^2} = \sqrt{(x - 0)^2 + (y - 0)^2 + (0 - 1)^2}$$

Squaring both sides:

$$x^2 + (y - 3)^2 + 4 = x^2 + y^2 + 1$$

Expanding:

$$x^2 + y^2 - 6y + 9 + 4 = x^2 + y^2 + 1$$

Simplifying:

$$-6y + 13 = 1$$

$$-6y = -12$$

$$y = 2$$

Setting $$AP = AQ$$:

$$\sqrt{(x - 0)^2 + (y - 3)^2 + (0 - 2)^2} = \sqrt{(x - 2)^2 + (y - 0)^2 + (0 - 3)^2}$$

Substituting $$y = 2$$:

$$\sqrt{x^2 + (2 - 3)^2 + 4} = \sqrt{(x - 2)^2 + 2^2 + 9}$$

$$\sqrt{x^2 + 1 + 4} = \sqrt{(x - 2)^2 + 13}$$

$$\sqrt{x^2 + 5} = \sqrt{(x - 2)^2 + 13}$$

Squaring both sides:

$$x^2 + 5 = (x - 2)^2 + 13$$

Expanding:

$$x^2 + 5 = x^2 - 4x + 4 + 13$$

Simplifying:

$$5 = -4x + 17$$

$$4x = 12$$

$$x = 3$$

Thus, $$A = (3, 2, 0)$$.

Given $$B = (1, 4, -1)$$ and $$C = (2, 0, -2)$$, we check the statements.

Statement (S1): $$\triangle ABC$$ is an isosceles right-angled triangle.

Compute the side lengths:

$$AB = \sqrt{(3 - 1)^2 + (2 - 4)^2 + (0 - (-1))^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$$

$$AC = \sqrt{(3 - 2)^2 + (2 - 0)^2 + (0 - (-2))^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$$

$$BC = \sqrt{(1 - 2)^2 + (4 - 0)^2 + (-1 - (-2))^2} = \sqrt{1 + 16 + 1} = \sqrt{18} = 3\sqrt{2}$$

Since $$AB = AC = 3$$, $$\triangle ABC$$ is isosceles.

Check Pythagoras theorem:

$$AB^2 + AC^2 = 3^2 + 3^2 = 9 + 9 = 18$$

$$BC^2 = (3\sqrt{2})^2 = 18$$

Since $$AB^2 + AC^2 = BC^2$$, $$\triangle ABC$$ is right-angled at $$A$$. Thus, (S1) is true.

Statement (S2): The area of $$\triangle ABC$$ is $$\frac{9\sqrt{2}}{2}$$.

Since $$\triangle ABC$$ is right-angled at $$A$$, area $$= \frac{1}{2} \times AB \times AC = \frac{1}{2} \times 3 \times 3 = \frac{9}{2}$$.

Alternatively, using vectors:

Vector $$\overrightarrow{AB} = B - A = (1 - 3, 4 - 2, -1 - 0) = (-2, 2, -1)$$

Vector $$\overrightarrow{AC} = C - A = (2 - 3, 0 - 2, -2 - 0) = (-1, -2, -2)$$

Cross product $$\overrightarrow{AB} \times \overrightarrow{AC}$$:

$$\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2 & 2 & -1 \\ -1 & -2 & -2 \end{vmatrix} = \mathbf{i}(2 \cdot (-2) - (-1) \cdot (-2)) - \mathbf{j}((-2) \cdot (-2) - (-1) \cdot (-1)) + \mathbf{k}((-2) \cdot (-2) - 2 \cdot (-1))$$

$$= \mathbf{i}(-4 - 2) - \mathbf{j}(4 - 1) + \mathbf{k}(4 + 2)$$

$$= -6\mathbf{i} - 3\mathbf{j} + 6\mathbf{k} = (-6, -3, 6)$$

Magnitude $$|\overrightarrow{AB} \times \overrightarrow{AC}| = \sqrt{(-6)^2 + (-3)^2 + 6^2} = \sqrt{36 + 9 + 36} = \sqrt{81} = 9$$

Area $$= \frac{1}{2} \times 9 = \frac{9}{2}$$

But $$\frac{9}{2} \neq \frac{9\sqrt{2}}{2}$$, so (S2) is false.

Therefore, only (S1) is true. The correct option is C.