$A(3,2,-1), B(4,1,1), C(6,2,5)$ and $D(3,3,3)$ are four points. $G_1, G_2, G_3$ and $G_4$ respectively are…
$A(3,2,-1), B(4,1,1), C(6,2,5)$ and $D(3,3,3)$ are four points. $G_1, G_2, G_3$ and $G_4$ respectively are the centroids of the triangles $\triangle B C D, \triangle C D A$, $\triangle D A B$ and $\triangle A B C$. The point of concurrence of the lines $A G_1, B G_2, C G_3$ and $D G_4$ is
(4, 2, 2)
(2, 4, 2)
(2, 2, 4)
(2, 2, 2)
Solution
Given points,
$A(3,2,-1), B(4,1,1), C(6,2,5)$ and $D(3,3,3)$, So,
$G_1$ is centroid of triangle BCD, $G_1 \equiv\left(\frac{13}{3}, \frac{6}{3}, \frac{9}{3}\right)$
$G_2$ is centroid of triangle CDA $G_2 \equiv\left(\frac{12}{3}, \frac{7}{3}, \frac{7}{3}\right)$
$\because$ The line $A G_1, B G_2, C G_3$ and $D G_4$ are concurrent, so point of concurrence of these four lines is point of intersection of lines $A G_1$ and $B G_2$.
Equation of line $A G_1$ is
$
\frac{x-3}{4 / 3}=\frac{y-2}{0}=\frac{z+1}{\frac{12}{3}}=r_1 \text { (let) }
$
So, point on line this $A G_1$ is $\left(3+\frac{4}{3} r_1, 2,-1+\frac{12}{3} r_1\right)$ and equation of line $B G_2$ is
$
\frac{x-4}{0}=\frac{y-1}{4 / 3}=\frac{z-1}{4 / 3}=r_2 \text { (let) }
$
So, point on line $B G_2$ is $\left(4,1+\frac{4}{3} r_2, 1+\frac{4}{3} r_2\right)$
Let the above point is the point of intersection, so
$
3+\frac{4}{3} r_1=4 \Rightarrow 2=1+\frac{4}{3} r_2
$
and $-1+\frac{12}{3} r_1=1+\frac{4}{3} r_2$, from these we are getting $r_1=\frac{3}{4}$ and $r_2=\frac{3}{4}$
So, required point of concurrence is $(4,2,2)$