Question

The vertices of $\triangle ABC$ are $A (2,0,0), B (0,1,0), C (0,0,2)$. Its orthocentre is $H$ and circumcentre is $S$. $P$ is a point equidistant from $A , B , C$ and the origin $O$.
$PA =$

• A. 1
• B. $\sqrt{2}$
• C. $\sqrt{\frac{3}{2}}$
• D. $\frac{3}{2}$

Answer 1

Answer: (D)

$P ( x , y , z ) \Rightarrow x ^2+ y ^2+ z ^2=( x -2)^2+ y ^2+ z ^2 \\
= x ^2+( y -1)^2+ z ^2= x ^2+ y ^2+( z -2)^2 $
$\Rightarrow x =1, y =\frac{1}{2}, z =1$
$ P \left(1, \frac{1}{2}, 1\right), A (2,0,0) \Rightarrow AP =\frac{3}{2}$