Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

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

In $xy$ plane $z=0$
So, $x^2-4x+4+y^2+9=x^2+y^2+1$
$\begin{array}{rl}& x=3\\ & 9+y^2-6y+9+4=x^2+y^2+1\end{array}$

So, $\mathrm{A}(3,2,0)$ also $\mathrm{B}(1,4,-1)\&\mathrm{C}(2,0,-2)$
Now $AB=\sqrt{4+4+1}=3$
$\begin{array}{rl}& \mathrm{AC}=\sqrt{1+4+4}=3\\ & \mathrm{BC}=\sqrt{1+16+1}=\sqrt{18}\end{array}$
$\mathrm{AB}=\mathrm{AC}$
isosceles $\Delta \&{\mathrm{AB}}^2+{\mathrm{AC}}^2={\mathrm{BC}}^2$
right angle $\Delta$
Area of $△\mathrm{ABC}=\frac{1}{2}\times$ base.height
$\frac{1}{2}\times 3\times 3=\frac{9}{2}$

So only $S_1$ is true