In Delta ABC,if A(2, 4, -1),B(4, 5, 1),and C( | vedclass

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(D) First,we calculate the vectors for the sides of the triangle: $\vec{AB} = (4-2, 5-4, 1-(-1)) = (2, 1, 2)$. The length is $|\vec{AB}| = \sqrt{2^2 +

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Isosceles right-angled

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(D) First,we calculate the vectors for the sides of the triangle: $\vec{AB} = (4-2, 5-4, 1-(-1)) = (2, 1, 2)$. The length is $|\vec{AB}| = \sqrt{2^2 + 1^2 + 2^2} = \sqrt{4+1+4} = \sqrt{9} = 3$. $\vec{BC} = (3-4, 6-5, -3-1) = (-1, 1, -4)$. The length is $|\vec{BC}| = \sqrt{(-1)^2 + 1^2 + (-4)^2} = \sqrt{1+1+16} = \sqrt{18} = 3\sqrt{2}$. $\vec{CA} = (2-3, 4-6, -1-(-3)) = (-1, -2, 2)$. The length is $|\vec{CA}| = \sqrt{(-1)^2 + (-2)^2 + 2^2} = \sqrt{1+4+4} = \sqrt{9} = 3$. Since $|\vec{AB}| = |\vec{CA}| = 3$,the triangle is isosceles. Now,check for the right-angle condition: $|\vec{AB}|^2 + |\vec{CA}|^2 = 3^2 + 3^2 = 9 + 9 = 18$. Also,$|\vec{BC}|^2 = (3\sqrt{2})^2 = 18$. Since $|\vec{AB}|^2 + |\vec{CA}|^2 = |\vec{BC}|^2$,the triangle is a right-angled triangle at vertex $A$. Therefore,$\Delta ABC$ is an isosceles right-angled triangle.

In $\Delta ABC$,if $A(2, 4, -1)$,$B(4, 5, 1)$,and $C(3, 6, -3)$,then $\Delta ABC$ is a ... triangle.

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