Concept:The equation of the plane whose intercepts are a, b, c on the x, y, z axes, respectively, is \(\displaystyle \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\) (a, b, c ? 0)Calculation:Since the plane \(\displaystyle \frac{2x}{k} + \frac{2y}{3} + \frac{z}{3} = 2\) pass through the point (2, 3, -6);So, it satisfies the given equation.So, \(\displaystyle \frac{2\times 2}{k} + \frac{2\times3}{3} + \frac{-6}{3} = 2\)? k = 2The given equation of the plane is \(\displaystyle \frac{2x}{k} + \frac{2y}{3} + \frac{z}{3} = 2\)Putting the value of k = 2, we get,\(\displaystyle \frac{2x}{2} + \frac{2y}{3} + \frac{z}{3} = 2\)? 3x + 2y + z = 6The equation of the plane in intercepts form,\(\displaystyle \frac{3x}{6}+\frac{2y}{6}+\frac{z}{6}=1??\)? \(\displaystyle \frac{x}{2}+\frac{y}{3}+\frac{z}{6}=1\)2, 3, and 6 are the intercepts made by the plane on the coordinate axes, respectively.So, Sum of the intercept = 2 + 3 + 6 = 11.? The value of (p + q + r) is 11.