The circumcentre of the triangle formed by th | vedclass

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Question

(C) Let the circumcentre be $P(x, y, z)$. The circumcentre is equidistant from the vertices $A, B$,and $C$. Thus,$PA^2 = PB^2 = PC^2$.$PA^2 = (x-3)^2

Vedclass · Accepted Answer

$(1,5,3)$

Vedclass · Answer

(C) Let the circumcentre be $P(x, y, z)$. The circumcentre is equidistant from the vertices $A, B$,and $C$. Thus,$PA^2 = PB^2 = PC^2$.$PA^2 = (x-3)^2 + (y-4)^2 + (z-5)^2$$PB^2 = (x-2)^2 + (y-3)^2 + (z-1)^2$$PC^2 = (x+1)^2 + (y-6)^2 + (z-1)^2$Equating $PB^2 = PC^2$:$(x-2)^2 + (y-3)^2 + (z-1)^2 = (x+1)^2 + (y-6)^2 + (z-1)^2$$x^2 - 4x + 4 + y^2 - 6y + 9 = x^2 + 2x + 1 + y^2 - 12y + 36$$-6x + 6y = 24 \implies -x + y = 4 \implies y = x + 4$.Equating $PA^2 = PB^2$:$(x-3)^2 + (y-4)^2 + (z-5)^2 = (x-2)^2 + (y-3)^2 + (z-1)^2$Substitute $y = x+4$:$(x-3)^2 + (x+4-4)^2 + (z-5)^2 = (x-2)^2 + (x+4-3)^2 + (z-1)^2$$(x-3)^2 + x^2 + (z-5)^2 = (x-2)^2 + (x+1)^2 + (z-1)^2$$x^2 - 6x + 9 + x^2 + z^2 - 10z + 25 = x^2 - 4x + 4 + x^2 + 2x + 1 + z^2 - 2z + 1$$-6x - 10z + 34 = -2x - 2z + 6$$-4x - 8z = -28 \implies x + 2z = 7 \implies z = \frac{7-x}{2}$.Since $P$ lies on the plane of the triangle,we check the options. For option $C(1, 5, 3)$:$y = 1+4 = 5$ (Matches).$z = (7-1)/2 = 3$ (Matches).Thus,the circumcentre is $(1, 5, 3)$.

Column $I$	Column $II$
$(A)$ The centroid of the triangle formed by $(2, 3, -1)$,$(5, 6, 3)$,$(2, -3, 1)$ is	$(p)$ $(2, 2, 2)$
$(B)$ The circumcentre of the triangle formed by $(1, 2, 3)$,$(2, 3, 1)$,$(3, 1, 2)$ is	$(q)$ $(3, 1, 4)$
$(C)$ The orthocentre of the triangle formed by $(2, 1, 5)$,$(3, 2, 3)$,$(4, 0, 4)$ is	$(r)$ $(1, 1, 0)$
$(D)$ The incentre of the triangle formed by $(0, 0, 0)$,$(3, 0, 0)$,$(0, 4, 0)$ is	$(s)$ $(3, 2, 1)$

The circumcentre of the triangle formed by the points $A(3,4,5)$,$B(2,3,1)$,and $C(-1,6,1)$ is:

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