If the vertices of a parallelogram $A B C D$ are $A(1,2,3), B(-1,-2,-1)$ and $C(2,3,2)$, then find the fourth vertex $D$.
$$\begin{aligned} &\text { Let the fourth vertex of the parallelogram } A B C D \text { is } D(x, y, z) \text {. Then, the mid-point of } A C \text { are }\\ &P\left(\frac{1+2}{2}, \frac{2+3}{2}, \frac{3+2}{2}\right) \text { i.e., } P\left(\frac{3}{2}, \frac{5}{2}, \frac{5}{2}\right) \text {. } \end{aligned}$$
$$\begin{aligned} &\text { Now, mid-point of } B D \text {, }\\ &\begin{aligned} & \frac{3}{2}=\frac{-1+x}{2} \Rightarrow x=4 \\ & \frac{5}{2}=\frac{-2+y}{2} \Rightarrow y=7 \\ & \frac{5}{2}=\frac{-1+z}{2} \Rightarrow z=6 \end{aligned} \end{aligned}$$
So, the coordinates of fourth vertex is (4, 7, 6).
Find the coordinate of the points which trisect the line segment joining the points $A(2,1,-3)$ and $B(5,-8,3)$.
Let the $P\left(x_1, y_1, z_1\right)$ and $Q\left(x_2, y_2, z_2\right)$ trisect line segment $A B$.
Since, the point $P$ divided line $A B$ in $1: 2$ internally, then
$$\begin{aligned} & x_1=\frac{2 \times 2+1 \times 5}{1+2}=\frac{9}{3}=3 \\ & y_1=\frac{2 \times 1+1 \times(-8)}{3}=\frac{-6}{3}=-2 \\ & z_1=\frac{2 \times(-3)+1 \times 3}{3}=\frac{-6+3}{3}=\frac{-3}{3}=-1 \end{aligned}$$
Since, the point $Q$ divide the line segment $A B$ in $2: 1$, then
$$\begin{aligned} & x_2=\frac{1 \times 2+2 \times 5}{3}=4 \\ & y_2=\frac{1 \times 1+(-8 \times 2)}{3}=-5 \\ & z_2=\frac{1 \times(-3)+2 \times 3}{3}=-1 \end{aligned}$$
So, the coordinates of $P$ are $(3,-2,-1)$ and the coordinates of $Q$ are $(4,-5,1)$.
If the origin is the centroid of a $\triangle A B C$ having vertices $A(a, 1,3)$, $B(-2, b,-5)$ and $C(4,7, c)$, then find the values of $a, b, c$.
Given that origin is the centroid of the $\triangle A B C$ i.e., $G(0,0,0)$.
$$\begin{array}{ll} \because \quad & 0=\frac{a-2+4}{3} \Rightarrow a=-2 \\ & 0=\frac{1+b+7}{3} \Rightarrow b=-8 \\ & 0=\frac{3-5+c}{3} \Rightarrow c=+2 \\ \therefore \quad & a=-2, b=-8 \text { and } c=2 \end{array}$$
If $A(2,2,-3), B(5,6,9), C(2,7,9)$ be the vertices of a triangle. The internal bisector of the angle $A$ meets $B C$ at the point $D$, then find the coordinates of $D$.
Let the coordinates of $D$ are $(x, y, z)$.
$$\begin{aligned} A B & =\sqrt{9+16+144}=\sqrt{169}=13 \\ A C & =\sqrt{0+25+144}=\sqrt{169}=\sqrt{13} \\ \Rightarrow \quad \frac{A B}{A C} & =\frac{13}{13} \Rightarrow A B=A C \\ \frac{B D}{D C} & =\frac{1}{1} \Rightarrow B D=D C \end{aligned}$$
Since, $D$ is divide the line $B C$ in two equal parts. So, $D$ is the mid-point of $B C$.
$$\begin{array}{ll} \therefore & x=\frac{5+2}{2}=7 / 2 \\ \Rightarrow & y=\frac{6+7}{2}=13 / 2 \\ \Rightarrow & z=\frac{9+9}{2}=9 \end{array}$$
So, the coordinates of $D$ are $\left(\frac{7}{2}, \frac{13}{2}, 9\right)$.
Show that the three points $A(2,3,4), B(1,2,-3)$ and $C(-4,1,-10)$ are collinear and find the ratio in which $C$ divides $A B$.
Given points are $A(2,3,4), B(-1,2,-3)$ and $C(-4,1,-10)$.
$$\begin{aligned} & \therefore \quad A B=\sqrt{(2+1)^2+(3-2)^2+(4+3)^2} \\ & =\sqrt{9+1+49}=\sqrt{59} \\ & B C=\sqrt{(-1+4)^2+(2-1)^2+(-3+10)^2} \\ & =\sqrt{9+1+49}=\sqrt{59} \\ & A C=\sqrt{(2+4)^2+(3-1)^2+(4+10)^2} \\ & =\sqrt{36+4+196} \\ & =\sqrt{236}=2 \sqrt{59} \\ & \text { Now, } \\ & A B+B C=\sqrt{59}+\sqrt{59}=2 \sqrt{59} \\ & \because \quad A B+B C=A C \end{aligned}$$
Hence, the points $A, B$ and $C$ are collinear.
Now, $$A C: B C=2 \sqrt{59}: \sqrt{59}=2: 1$$
So, $C$ divide $A B$ in $2: 1$ externally.