The incenter and centroid of the triangle,who | vedclass

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(D) Given vertices are $A \equiv(0,3,0), B \equiv(0,0,4)$,and $C \equiv(0,3,4)$.First,calculate the lengths of the sides:$a = BC = \sqrt{(0-0)^2 + (3-

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$(0,2,3),\left(0,2, \frac{8}{3}\right)$

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(D) Given vertices are $A \equiv(0,3,0), B \equiv(0,0,4)$,and $C \equiv(0,3,4)$.First,calculate the lengths of the sides:$a = BC = \sqrt{(0-0)^2 + (3-0)^2 + (4-4)^2} = \sqrt{0+9+0} = 3$$b = CA = \sqrt{(0-0)^2 + (3-3)^2 + (4-0)^2} = \sqrt{0+0+16} = 4$$c = AB = \sqrt{(0-0)^2 + (0-3)^2 + (4-0)^2} = \sqrt{0+9+16} = \sqrt{25} = 5$Now,the Incenter $I$ is given by $\left(\frac{ax_1+bx_2+cx_3}{a+b+c}, \frac{ay_1+by_2+cy_3}{a+b+c}, \frac{az_1+bz_2+cz_3}{a+b+c}\right)$:$I = \left(\frac{3(0)+4(0)+5(0)}{3+4+5}, \frac{3(3)+4(0)+5(3)}{3+4+5}, \frac{3(0)+4(4)+5(4)}{3+4+5}\right)$$I = \left(\frac{0}{12}, \frac{9+0+15}{12}, \frac{0+16+20}{12}\right) = \left(0, \frac{24}{12}, \frac{36}{12}\right) = (0, 2, 3)$The Centroid $G$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$:$G = \left(\frac{0+0+0}{3}, \frac{3+0+3}{3}, \frac{0+4+4}{3}\right) = \left(0, \frac{6}{3}, \frac{8}{3}\right) = \left(0, 2, \frac{8}{3}\right)$Thus,the incenter and centroid are $(0, 2, 3)$ and $\left(0, 2, \frac{8}{3}\right)$ respectively.

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 incenter and centroid of the triangle,whose vertices are $A \equiv(0,3,0), B \equiv(0,0,4)$,and $C \equiv(0,3,4)$,are respectively given by

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