The intercepts cut off by the planes \(2x+y-z=5\) are

\(2,1,-1\) \(\frac{5}{2},5,-5\) \(5,\frac{5}{2},-5\) \(5,-5,\frac{5}{2}\)

\(\frac{5}{2},5,-5\)

Given \(2x+y-z=5\) $\Rightarrow \frac{2x}{5} + \frac{y}{5} - \frac{z}{5} = 1$ $ \Rightarrow \frac{x}{5/2} + \frac{y}{5} - \frac{z}{5} = 1$ Compare this equation with standard form $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ $ \Rightarrow a = \frac{5}{2} , b = 5 , c = -5$