JEE MAIN - Mathematics (2025 - 28th January Morning Shift - No. 1)
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 $\frac{9 \sqrt{2}}{2}$,
Explanation
$$\begin{aligned} & A(x, y, z) \text { Let } P(0,3,2), Q(2,0,3), R(0,0,1) \\ & A P=A Q=A R \\ & x^2+(y-3)^2+(z-2)^2=(x-2)^2+y^2+(z-3)^2=x^2+ \\ & y^2+(z-1)^2 \end{aligned}$$
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 $A B=\sqrt{4+4+1}=3$
$$\begin{aligned} &\begin{aligned} & \mathrm{AC}=\sqrt{1+4+4}=3 \\ & \mathrm{BC}=\sqrt{1+16+1}=\sqrt{18} \\ & \mathrm{AB}=\mathrm{AC} \\ & \text { isosceles } \Delta \& \mathrm{AB}^2+\mathrm{AC}^2=\mathrm{BC}^2 \\ & \text { right angle } \Delta \\ & \text { Area of } \triangle \mathrm{ABC}=\frac{1}{2} \times \text { base.height } \\ & \frac{1}{2} \times 3 \times 3=\frac{9}{2} \end{aligned}\\ &\text { So only } S_1 \text { is true } \end{aligned}$$
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