Sum of the intercepts cut off by the plane $2x+y-z=5$ is !*n1*!

The correct answer is $\dfrac{5}{2}$

The intercept equation of a plane with $x$, $y$ and $z$ intercepts $a$, $b$ and $c$ respectively, is given by

$\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1$ ----(1)

Given equation is $2x+y-z=5$ ⇒ $\dfrac{2x}{5}+ \dfrac{y}{5}-\dfrac{z}{5}=1$ ----(2)

Comparing equation (2) with equation (1), we get,

$a = \dfrac{5}{2}$, $b=5$ and $c=-5$

∴ Sum of the intercepts = $\dfrac{5}{2}+5-5=\dfrac{5}{2}$