Find the intercepts cut off by the plane

$2 x+y-z=5$                               $\ldots(1)$

Dividing both sides of equation (1) by 5, we obtain

$\frac{2}{5} x+\frac{y}{5}-\frac{z}{5}=1$

$\Rightarrow \frac{x}{\frac{5}{2}}+\frac{y}{5}+\frac{z}{-5}=1$            $\ldots(2)$

It is known that the equation of a plane in intercept form is $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$, where $a, b, c$ are the intercepts cut off by the plane at $x, y$, and $z$ axes respectively.

Therefore, for the given equation,

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

Thus, the intercepts cut off by the plane are $\frac{5}{2}, 5$, and $-5$.