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

Finding the distance $AB,BC,AC$

$A(3,6,9),B(10,20,30),C(25,-415)$
Using distance formula
$AB=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}$

Here, $x_1=3,y_1=6,z_1=9$

$x_2=10,y_2=20,z_2=30$

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

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

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

$=\sqrt{686}$

Calculating $BC$

$B(10,20,30),$
$C(25,-41,5)$

$BC=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}$

Here, $x_1=10,y_1=20,z_1=30$
$x_2=25,y_2=-41,z_2=5$

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

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

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

$=\sqrt{4571}$

$\therefore AB=\sqrt{686}\text{and}BC=\sqrt{4571}$

Calculating $AC$

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

$\therefore AB=\sqrt{686}\text{and}BC=\sqrt{4571}$

$AC=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}$

Here, $x_1=3,y_1=6,z_1=9$
$x_2=25,y_2=-41,z_2=5$

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

$=\sqrt{(-22{)}^2+(6+41{)}^2+(4{)}^2}$

$=\sqrt{484+(47{)}^2+16}$

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

$=\sqrt{2709}$

Now,

$AB=\sqrt{686},BC=\sqrt{4571},AC=\sqrt{2709}$

Using Pythagoras Theorem In Right angle triangle,
$(\text{Hypotenuse}{)}^2=(\text{Height}{)}^2+(\text{Base}{)}^2$

we take Hypotenuse =$\sqrt{4571}$
As,
$A{B}^2+A{C}^2\ne B{C}^2$
So, the triangle formed is not right angles triangle.