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Q.
The points $A (1,2,3), B (-1,-2,-3)$ and $C (2,3,2)$ are
three vertices of a parallelogram $ABCD$. The equation of $CD$ is
Three Dimensional Geometry
Solution:
Given $A (1,2,3), B (-1,-2,-1)$ and $C (2,3,2)$, Let $D$ be $(\alpha, \beta, \gamma) .$ Since $ABCD$ is a parallelogram, diagonals $AC$ and BD bisect each other i.e., mid-point of segment $AC$ is same as mid-point of segment $BD$.
$\Rightarrow \left(\frac{1+2}{2}, \frac{2+3}{2}, \frac{3+2}{2}\right)=\left(\frac{\alpha-1}{2}, \frac{\beta-2}{2}, \frac{\gamma-1}{2}\right)$
$\Rightarrow \alpha-1=3, \beta-2=5, \gamma-1=5$
$\Rightarrow \alpha=4, \beta=7, \gamma=6$
Hence, the point D is $(4,7,6)$.
We have $C(2,3,2)$ and $D(4,7,6)$
$\therefore $ Equation of line $CD$ is
$\frac{x-2}{4-2}=\frac{y-3}{7-3}=\frac{z-2}{6-2}$
$\Rightarrow \frac{x-2}{2}=\frac{y-3}{4}=\frac{z-2}{4}$
i.e., $\frac{x-2}{1}=\frac{y-3}{2}=\frac{z-2}{2}$
is the required equation of line.