Let $\mathrm{A}(x, y, z)$ be a point in $x y$-plane, which is equidistant from three points $(0,3,2),(2,0,3)$ and $(0,0,1)$.

Let $\mathrm{B}=(1,4,-1)$ and $\mathrm{C}=(2,0,-2)$. Then among the statements

(S1) : $\triangle \mathrm{ABC}$ is an isosceles right angled triangle, and

(S2) : the area of $\triangle \mathrm{ABC}$ is $\frac{9 \sqrt{2}}{2}$,

<p>$$\begin{aligned} & A(x, y, z) \text { Let } P(0,3,2), Q(2,0,3), R(0,0,1) \\ & A P=A Q=A R \\ & x^2+(y-3)^2+(z-2)^2=(x-2)^2+y^2+(z-3)^2=x^2+ \\ & y^2+(z-1)^2 \end{aligned}$$</p> <p>In xy plane $\mathrm{z}=0$</p> <p>So, $x^2-4 x+4+y^2+9=x^2+y^2+1$</p> <p>$$\begin{aligned} & \Rightarrow y=2 \\ & x=3 \\ & 9+y^2-6 y+9+4=x^2+y^2+1 \end{aligned}$$</p> <p>So, $A(3,2,0)$ also $B(1,4,-1) \& C(2,0,-2)$</p> <p>Now $A B=\sqrt{4+4+1}=3$</p> <p>$$\begin{aligned} &\begin{aligned} & \mathrm{AC}=\sqrt{1+4+4}=3 \\ & \mathrm{BC}=\sqrt{1+16+1}=\sqrt{18} \\ & \mathrm{AB}=\mathrm{AC} \\ & \text { isosceles } \Delta \& \mathrm{AB}^2+\mathrm{AC}^2=\mathrm{BC}^2 \\ & \text { right angle } \Delta \\ & \text { Area of } \triangle \mathrm{ABC}=\frac{1}{2} \times \text { base.height } \\ & \frac{1}{2} \times 3 \times 3=\frac{9}{2} \end{aligned}\\ &\text { So only } S_1 \text { is true } \end{aligned}$$</p>