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

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$\frac{x}{1} = \frac{y}{2} = \frac{z}{2}$

$\frac{x + 2}{1} = \frac{y + 3}{2} = \frac{z - 2}{2}$

$\frac{x}{2} = \frac{y}{3} = \frac{z}{2}$

$\frac{x - 2}{1} = \frac{y - 3}{2} = \frac{z - 2}{2}$

answer is D.

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Detailed Solution

Given $A (1, 2, 3), B (- 1, - 2, - 1) and C (2, 3, 2)$ , Let D be $(α, β, γ)$ , Since ABCD is a parallelogram, diagonals Since ABCD is a parallelogram, diagonals

$\begin{array}{l} \Rightarrow (\frac{1 + 2}{2}, \frac{2 + 3}{2}, \frac{3 + 2}{2}) = (\frac{α - 1}{2}, \frac{β - 2}{2}, \frac{γ - 1}{2}) \\ \Rightarrow α - 1 = 3, β - 2 = 5, γ - 1 = 5 \\ \Rightarrow α = 4, β = 7, γ = 6 \end{array}$

Hence, the point D is (4, 7, 6).

We have C(2, 3, 2) and D(4, 7, 6).

$∴$ Equation of line CD is

$\begin{array}{l} \frac{x - 2}{4 - 2} = \frac{y - 3}{7 - 3} = \frac{z - 2}{6 - 2} \\ \Rightarrow \frac{x - 2}{2} = \frac{y - 3}{4} = \frac{z - 2}{4} \end{array}$