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

Here, A(3, 6, 9), B(10, 20, 30) and C(25, -41, 5) are vertices of

$△ABC$

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

=$\sqrt{(7{)}^2+(14{)}^2+(21{)}^2}$

=$\sqrt{49+196+441}$

=$\sqrt{686}=7\sqrt{14}$

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

=$\sqrt{(15{)}^2+(-61{)}^2+(-25{)}^2}$

=$\sqrt{225+3721+625}$

=$\sqrt{4571}$

CA=$\sqrt{(3-25{)}^2+(6-41{)}^2+(9-5{)}^2}$

=$\sqrt{(-22{)}^2+(47{)}^2+(-4{)}^2}$

=$\sqrt{484+2209+16}$

=$\sqrt{2709}=3\sqrt{301}$

$A{B}^2+B{C}^2={\left(7\sqrt{14}\right)}^2+{\left(\sqrt{4571}\right)}^2$

686 + 4571 = 5257

$C{A}^2$ = 2709

$\therefore A{B}^2+B{C}^2\ne C{A}^2$

$△ABC$ is right angled at B if $C{A}^2=A{B}^2+B{C}^2+B{C}^2$ but $C{A}^2\ne A{B}^2+B{C}^2$

Hence, the points are not vertices of a right-angled triangle.