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 $\frac{9 \sqrt{2}}{2}$,

Question

Aicharya · Accepted Answer

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

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

Solution