If A(1,2,3), B(2,-3,1), C(3,2,-1) are three v | vedclass

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(D) Let the vertices of the tetrahedron be $A(1,2,3), B(2,-3,1), C(3,2,-1)$ and $D(a, b, c)$.Since $G\left(\frac{5}{2}, \frac{3}{2}, \frac{9}{4}\right

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$\left(3, \frac{8}{3}, \frac{7}{2}\right)$

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(D) Let the vertices of the tetrahedron be $A(1,2,3), B(2,-3,1), C(3,2,-1)$ and $D(a, b, c)$.Since $G\left(\frac{5}{2}, \frac{3}{2}, \frac{9}{4}\right)$ is the centroid of the tetrahedron $ABCD$, we have:$G = \frac{A+B+C+D}{4}$$\left(\frac{5}{2}, \frac{3}{2}, \frac{9}{4}\right) = \frac{(1+2+3+a, 2-3+2+b, 3+1-1+c)}{4}$$\left(\frac{5}{2}, \frac{3}{2}, \frac{9}{4}\right) = \frac{(6+a, 1+b, 3+c)}{4}$Multiplying by $4$, we get:$(10, 6, 9) = (6+a, 1+b, 3+c)$Comparing the coordinates:$6+a = 10 \Rightarrow a = 4$$1+b = 6 \Rightarrow b = 5$$3+c = 9 \Rightarrow c = 6$Thus, $D = (4, 5, 6)$.Now, we find the point $P$ that divides $GD$ in the ratio $1:2$ using the section formula:$P = \left(\frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n}, \frac{m z_2 + n z_1}{m+n}\right)$Here $m=1, n=2$, $G = \left(\frac{5}{2}, \frac{3}{2}, \frac{9}{4}\right)$ and $D = (4, 5, 6)$.$x = \frac{1(4) + 2(5/2)}{1+2} = \frac{4+5}{3} = \frac{9}{3} = 3$$y = \frac{1(5) + 2(3/2)}{1+2} = \frac{5+3}{3} = \frac{8}{3}$$z = \frac{1(6) + 2(9/4)}{1+2} = \frac{6+9/2}{3} = \frac{21/2}{3} = \frac{7}{2}$Therefore, the required point is $\left(3, \frac{8}{3}, \frac{7}{2}\right)$.

If $A(1,2,3), B(2,-3,1), C(3,2,-1)$ are three vertices of a tetrahedron $ABCD$ and $G\left(\frac{5}{2}, \frac{3}{2}, \frac{9}{4}\right)$ is its centroid, then the point which divides $GD$ in the ratio $1:2$ is

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