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(A) The centroid $G$ of a triangle with vertices $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,and $(x_3, y_3, z_3)$ is given by $(\frac{x_1+x_2+x_3}{3}, \frac{

Vedclass · Accepted Answer

$A-s, B-p, C-q, D-r$

Vedclass · Answer

(A) The centroid $G$ of a triangle with vertices $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,and $(x_3, y_3, z_3)$ is given by $(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3})$.For vertices $(2, 3, -1)$,$(5, 6, 3)$,$(2, -3, 1)$:$G = (\frac{2+5+2}{3}, \frac{3+6-3}{3}, \frac{-1+3+1}{3}) = (\frac{9}{3}, \frac{6}{3}, \frac{3}{3}) = (3, 2, 1)$. Thus,$A-s$.$(B)$ For an equilateral triangle,the circumcentre is the same as the centroid.For vertices $(1, 2, 3)$,$(2, 3, 1)$,$(3, 1, 2)$:$C = (\frac{1+2+3}{3}, \frac{2+3+1}{3}, \frac{3+1+2}{3}) = (\frac{6}{3}, \frac{6}{3}, \frac{6}{3}) = (2, 2, 2)$. Thus,$B-p$.$(C)$ For an equilateral triangle,the orthocentre is the same as the centroid.For vertices $(2, 1, 5)$,$(3, 2, 3)$,$(4, 0, 4)$:$O = (\frac{2+3+4}{3}, \frac{1+2+0}{3}, \frac{5+3+4}{3}) = (\frac{9}{3}, \frac{3}{3}, \frac{12}{3}) = (3, 1, 4)$. Thus,$C-q$.$(D)$ For a right-angled triangle with vertices at $(0, 0, 0)$,$(a, 0, 0)$,and $(0, b, 0)$,the incentre is $(\frac{ab}{a+b+\sqrt{a^2+b^2}}, \frac{ab}{a+b+\sqrt{a^2+b^2}}, 0)$.Here,vertices are $(0, 0, 0)$,$(3, 0, 0)$,$(0, 4, 0)$. So,$a=3, b=4$. The hypotenuse $c = \sqrt{3^2+4^2} = 5$.$I = (\frac{3 	imes 4}{3+4+5}, \frac{3 	imes 4}{3+4+5}, 0) = (\frac{12}{12}, \frac{12}{12}, 0) = (1, 1, 0)$. Thus,$D-r$.

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	$(s)$ $(3, 2, 1)$

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