The circumcentre of the triangle formed by th | vedclass

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(D) Let $A(1, 2, 3), B(3, -1, 5), C(4, 0, -3)$ be the vertices of the triangle. Let $O(x, y, z)$ be the circumcentre. Then $OA = OB = OC$,which implie

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$\left(\frac{7}{2}, -\frac{1}{2}, 1\right)$

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(D) Let $A(1, 2, 3), B(3, -1, 5), C(4, 0, -3)$ be the vertices of the triangle. Let $O(x, y, z)$ be the circumcentre. Then $OA = OB = OC$,which implies $OA^2 = OB^2 = OC^2$.$OA^2 = OB^2 \Rightarrow (x-1)^2 + (y-2)^2 + (z-3)^2 = (x-3)^2 + (y+1)^2 + (z-5)^2$$x^2 - 2x + 1 + y^2 - 4y + 4 + z^2 - 6z + 9 = x^2 - 6x + 9 + y^2 + 2y + 1 + z^2 - 10z + 25$$4x - 6y + 4z = 20 \Rightarrow 2x - 3y + 2z = 10 \quad \dots (i)$$OB^2 = OC^2 \Rightarrow (x-3)^2 + (y+1)^2 + (z-5)^2 = (x-4)^2 + y^2 + (z+3)^2$$x^2 - 6x + 9 + y^2 + 2y + 1 + z^2 - 10z + 25 = x^2 - 8x + 16 + y^2 + z^2 + 6z + 9$$2x + 2y - 16z = -10 \Rightarrow x + y - 8z = -5 \quad \dots (ii)$$OA^2 = OC^2 \Rightarrow (x-1)^2 + (y-2)^2 + (z-3)^2 = (x-4)^2 + y^2 + (z+3)^2$$x^2 - 2x + 1 + y^2 - 4y + 4 + z^2 - 6z + 9 = x^2 - 8x + 16 + y^2 + z^2 + 6z + 9$$6x - 4y - 12z = 11 \quad \dots (iii)$Solving equations $(i), (ii),$ and $(iii)$,we get $x = \frac{7}{2}, y = -\frac{1}{2}, z = 1$.Thus,the circumcentre is $\left(\frac{7}{2}, -\frac{1}{2}, 1\right)$.

The circumcentre of the triangle formed by the points $(1, 2, 3), (3, -1, 5), (4, 0, -3)$ is

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