If p, q and r are the intercepts made by the plane on the coordinate axes respectively, then what is (p + q + r) equal to ?

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 = 2

The 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 = 6

The 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.