Sum of x, y and z intercepts of a plane passing through three points (1, 0, 0), (0, 2, 3) and (0, 0, 5) is:

11

The correct answer is Option (1) - 11

x intercept is (1, 0, 0)

z intercept is (0, 0, 5)

let y intercept be (0, β, 0)

so eq → $\frac{x}{1}+\frac{y}{β}+\frac{z}{5}=1$

(0, 2, 3) falls on plane

$0+\frac{2}{β}+\frac{3}{5}=1⇒β=5$

so sum of intercepts = 1 + 5 + 5 = 11