Let 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

 B = (1, 4, –1) and C = (2, 0, –2). Then among the statements 

(S1) : $∆$ABC is an isosceles right angled triangle and 

(S2) : the area of $∆$ABC is $\frac{9\sqrt{2}}{2}$ .

$\begin{array}{l}A(x,y,z)\text{ Let }P(0,3,2),Q(2,0,3),R(0,0,1)\\ AP=AQ=AR\\ x^2+(y-3{)}^2+(z-2{)}^2=(x-2{)}^2+y^2+(z-3{)}^2=x^2+\\ y^2+(z-1{)}^2\end{array}$

In xy plane z = 0

So, $x^2-4x+4+y^2+9=x^2+y^2+1$

So, A(3,2,0) also B(1,4,–1) & C(2,0,–2)
Now 

$\begin{array}{l}\mathrm{AB}=\sqrt{4+4+1}=3\\ \mathrm{AC}=\sqrt{1+4+4}=3\\ \mathrm{BC}=\sqrt{1+16+1}=\sqrt{18}\end{array}$

AB = AC
isosceles $\mathrm{\Delta }\mathrm{\&}{\mathrm{AB}}^2+{\mathrm{AC}}^2={\mathrm{BC}}^2$

right angle $∆$$\mathrm{ABC}=\frac{1}{2}\times$base.height

$\frac{1}{2}\times 3\times 3=\frac{9}{2}$

So only S₁ is true