Are the points A(3, 6, 9),B(10, 20, 30),and C | vedclass

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(B) Using the distance formula $d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2$,we calculate the squares of the sides:$AB^2 = (10 - 3)^2 + (20 -

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(B) Using the distance formula $d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2$,we calculate the squares of the sides:$AB^2 = (10 - 3)^2 + (20 - 6)^2 + (30 - 9)^2 = 7^2 + 14^2 + 21^2 = 49 + 196 + 441 = 686$$BC^2 = (25 - 10)^2 + (-41 - 20)^2 + (5 - 30)^2 = 15^2 + (-61)^2 + (-25)^2 = 225 + 3721 + 625 = 4571$$CA^2 = (3 - 25)^2 + (6 - (-41))^2 + (9 - 5)^2 = (-22)^2 + 47^2 + 4^2 = 484 + 2209 + 16 = 2709$For a right-angled triangle,the sum of the squares of two sides must equal the square of the third side (Pythagoras theorem).Checking the sums:$AB^2 + CA^2 = 686 + 2709 = 3395 
eq 4571 (BC^2)$$AB^2 + BC^2 = 686 + 4571 = 5257 
eq 2709 (CA^2)$$BC^2 + CA^2 = 4571 + 2709 = 7280 
eq 686 (AB^2)$Since none of the combinations satisfy the condition $a^2 + b^2 = c^2$,the points $A, B, C$ do not form a right-angled triangle.

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

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