Given plane is 4x + 3y – 2z = – 2
Dividing  with ‘–2’ on both sides
$\begin{array}{l}\Rightarrow \frac{4\mathrm{x}}{-2}+\frac{3\mathrm{y}}{-2}-\frac{2\mathrm{z}}{-2}=\frac{-2}{-2}\\ \Rightarrow -2\mathrm{x}+\frac{3\mathrm{y}}{-2}+\mathrm{z}=1\\ \Rightarrow \frac{\mathrm{x}}{\left(-\frac{1}{2}\right)}+\frac{\mathrm{y}}{\left(-\frac{2}{3}\right)}+\frac{\mathrm{z}}{1}=1\end{array}$
Which is of the form $\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}+\frac{\mathrm{z}}{\mathrm{c}}=1$
where $\mathrm{x}\text{-intercept }=\mathrm{a}=-\frac{1}{2},\mathrm{y}\text{-intercept }=\mathrm{b}=-\frac{2}{3}$
and z-intercept = c = 1