Q.The equation of the plane whose intercepts on the $x$, $y$ and $z$-axes are $-2$, $3$ and $4$ respectively, will be

Use the intercept form, then clear denominators.

Intercepts $-2, 3, 4$ give

$\frac{x}{-2} + \frac{y}{3} + \frac{z}{4} = 1.$

Multiply through by $12$: …

The $x$-intercept is found by putting $y=z=0$: $3x=13\Rightarrow x=\dfrac{13}{3}$.

The intercept on the $x$-axis is the $x$-value where the plane meets that axis, i.e. where $y=0$ and $z=0$.

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Writing the plane in intercept form and using the centroid formula gives intercepts $3a,3b,3c$.

Let the plane meet the axes at $A=(p,0,0)$, $B=(0,q,0)$, $C=(0,0,r)$, so its equation in intercept form is $\dfrac{x}{p}+\dfrac{y}{q}+\dfrac{z}{r}=1$.

The centroid of triangle $ABC$ is

$\left(\dfrac{p+0+0}{3},\dfrac{0+q+0}{3},\dfrac{0+0+r}{3}\right) = \left(\dfrac{p}{3},\dfrac{q}{3},\dfrac{r}{3}\right)$

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