Question
Write the intercepts made by the plane 2x − 3y + 4z = 12 on the coordinate axes.

Answer

The given equation of the plane is
2x − 3y + 4z = 12
Dividing both sides by 12, we get
$\Rightarrow\frac{2\text{x}}{\text{12}}+\frac{-3\text{y}}{\text{12}}+\frac{4\text{z}}{\text{12}}=\frac{12}{12}$
$\Rightarrow\frac{\text{x}}{\text{6}}+\frac{\text{y}}{-\text{4}}+\frac{\text{z}}{\text{3}}=1\ ....(1)$
We know that the equation of the plane whose intercepts on the coordianate axes are a, b and c is
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1\ ....(2)$
Comparing (1) and (2), we get
a = 6, b = -4 and c = 3.

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