[tex]A(-1,3,4) , B(3,-1,1,),C(5,1,1)
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\\d(A,B)= \sqrt{(-1-3)^2+(3-(-1))^2+(4-1)^2}= \sqrt{16+16+9}= \sqrt{41}
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\\d(B,C)= \sqrt{(3-5)^2+(-1-1)^2+(1-1)^2}= \sqrt{4+4+0}= \sqrt{8}
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\\d(A,C)= \sqrt{(-1-5)^2+(3-1)^2+(4-1)^2}= \sqrt{36+4+9} = \sqrt{49}
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\\AC^2=AB^2+BC^2
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\\ \sqrt{49}^2=\sqrt{41}^2+\sqrt{8}^2
\\
\\49 =41+8
\\
\\49=49[/tex]
ABC is a right triangle.