Find the intercepts cut off by the plane 2x + y – z = 5.

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Given plane is 2x + y – z = 5 ……(i)

Dividing both sides of the equation (i) by 5,

(⅖)x + (y/5) – (z/5) = 1

x/(5/2) + y/5 – z/5 =1

We know that,

The equation of a plane in intercept form is (x/a) + (y/b) + (z/c) = 1, where a, b, c are intercepts cut off by the plane at x, y, z-axes respectively.

For the given equation,

a = 5/2, b = 5, c = -5

Hence, the intercepts cut off by the plane are 5/2, 5 and -5.

Thanks