Let $\mathbf{A}(x, y, z)$ be a point in $xy$ -plane, which is equidistant from three points $(0, 3, 2)$, $(2, 0, 3)$ and $(0, 0, 1)$. Let $\mathbf{B} = (1, 4, -1)$ and $\mathbf{C} = (2, 0, -2)$. Then among the statements (S1): $\triangle ABC$ is an isosceles right angled triangle, and (S2): the area of $\triangle ABC$ is $\frac{9 \sqrt{2}}{2}$.

Step 1: Find the coordinates of point A

Since point A lies in the $xy$ -plane, its $z$ -coordinate is 0. Let A be $(x, y, 0)$. We are given that A is equidistant from three points: $P_1(0, 3, 2)$, $P_2(2, 0, 3)$, and $P_3(0, 0, 1)$. We use the distance formula to set up equations where $AP_1 = AP_2 = AP_3$.

Step 2: Solve for x and y

We equate the squared distances to eliminate the square roots. First, we equate $AP_1^2 = AP_2^2$ to get a relationship between $x$ and $y$. Then, we equate $AP_1^2 = AP_3^2$ to solve for $y$. Finally, we substitute the value of $y$ back into the first relationship to find $x$. This gives us the coordinates of point A.

Step 3: Calculate side lengths of triangle ABC

Now that we have the coordinates of A, B, and C, we can calculate the lengths of the sides of $\triangle ABC$ using the distance formula. We find the lengths of AB, BC, and AC.

Step 4: Verify statement (S1): Isosceles right-angled triangle

We observe that $AB = AC = 3$, which means $\triangle ABC$ is an isosceles triangle. To check if it's a right-angled triangle, we apply the Pythagorean theorem. Since $AB^2 + AC^2 = BC^2$, the triangle is right-angled at A. Therefore, statement (S1) is true.

Step 5: Verify statement (S2): Area of triangle ABC

Since $\triangle ABC$ is a right-angled triangle with the right angle at A, we can use AB and AC as the base and height. The area of a right-angled triangle is half the product of its perpendicular sides. We calculate the area and compare it with the given value in statement (S2). Since $\frac{9}{2} \neq \frac{9\sqrt{2}}{2}$, statement (S2) is false.

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