Are the points $A(3,6,9),B(10,20,30)$ and $C(25,-41,5)$, vertices of a right angled triangle?

The correct option is B False

$A(3,6,9)B(10,20,30)C(25,-41,5)$

$AB=\sqrt{(3-10{)}^2+(6-20{)}^2+(9-30{)}^2}=\sqrt{686}$

$BC=\sqrt{(10-25{)}^2+(20+41{)}^2+(30-5{)}^2}=\sqrt{4571}$

$CA=\sqrt{(35-3{)}^2+(-41-6{)}^2+(9-5{)}^2}=\sqrt{3249}$

$BC>AB>\&>CA$

Hence $B{C}^2=A{B}^2+C{A}^2$ for right angle $△$ but

$B{C}^2=4571$

$A{B}^2+C{A}^2=3935$

$B{C}^2\ne A{B}^2+C{A}^2$ Hence in right angle $△$