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:
Answer & explanation
Correct answer: option 1
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