QJEEM25Mab28a: JEE Main 2025 Mathematics - Class 12 JEEM25Mab28a Solution: Let be a point in -plane, which is equidistant from three…

$\begin{aligned}& A(x, y, z) \text { Let } P(0,3,2), Q(2,0,3), R(0,0,1) \\& A P=A Q=A R \\& x^2+(y-3)^2+(z-2)^2=(x-2)^2+y^2+(z-3)^2=x^2+ \\& y^2+(z-1)^2\end{aligned}$

In xy plane $\mathrm{z}=0$

So, $x^2-4 x+4+y^2+9=x^2+y^2+1$

$\begin{aligned}& \Rightarrow y=2 \\& x=3 \\& 9+y^2-6 y+9+4=x^2+y^2+1\end{aligned}$

So, $A(3,2,0)$ also $B(1,4,-1) \& C(2,0,-2)$

Now $A B=\sqrt{4+4+1}=3$

$\begin{aligned}&\begin{aligned}& \mathrm{AC}=\sqrt{1+4+4}=3 \\& \mathrm{BC}=\sqrt{1+16+1}=\sqrt{18} \\& \mathrm{AB}=\mathrm{AC} \\& \text { isosceles } \Delta \& \mathrm{AB}^2+\mathrm{AC}^2=\mathrm{BC}^2 \\& \text { right angle } \Delta \\& \text { Area of } \triangle \mathrm{ABC}=\frac{1}{2} \times \text { base.height } \\& \frac{1}{2} \times 3 \times 3=\frac{9}{2}\end{aligned}\\&\text { So only } S_1 \text { is true }\end{aligned}$