Concept:

1. Let A (x1, y1), B (x2, y2) and C (x3, y3) be the vertices of a Δ ABC, then area of Δ ABC = \(\frac{1}{2} \cdot \left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1\\ {{x_2}}&{{y_2}}&1\\ {{x_3}}&{{y_3}}&1 \end{array}} \right|\)

2. Distance formula:  A(x, y) and B(a, b) be any two points, by distance formula,

 \(\rm AB = \sqrt {(x - a)^2 + (y - b)^2}\)

3. (a + b)2 = a2 + 2ab + b2

4. (a - b)2 = a2 - 2ab + b2

Calculation:

Given that, coordinates of point A, C and D are (3, 2), (- 3, 2) and (k, -1) respectively. 

Area of Δ ABC = 9 sq. unit

According to the question, |AD| = |CD| 

Using distance formula,

\(\sqrt {(3 - k)^2 + (2 + 1)^2}\ =\ \sqrt {(-3 \ -\ k)^2 + (2\ +\ 1)^2}\)

Taking square of both side

⇒ (3 - k)² + 9 = (3 + k)2 + 9

⇒ 3 + k² - 6k = 9 + k² + 6k

⇒ 12k = 0 ⇒ k = 0