Answer
We show that the $z$-coordinate of the centroid is $\bar z = \frac{c}{4}$.
By symmetry, the coordinates of the centroid: $\left( {\bar x,\bar y,\bar z} \right) = \left( {\frac{a}{4},\frac{b}{4},\frac{c}{4}} \right)$.
Work Step by Step
We have the region ${\cal W}$, the tetrahedron bounded by the coordinate planes and the plane
$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$
as is shown in Figure 14.
Multiplying both sides by $c$ gives $\frac{c}{a}x + \frac{c}{b}y + z = c$.
So, $z = c - \frac{c}{a}x - \frac{c}{b}y$.
We can consider ${\cal W}$ as a $z$-simple region lying over the triangle ${\cal D}$ in the $xy$-plane bounded by $x=0$, $y=0$, and the line $y = b - \frac{b}{a}x$.
Thus, ${\cal W}$ is the region between the planes $z=0$ and $z = c - \frac{c}{a}x - \frac{c}{b}y$. The projection of ${\cal W}$ onto the $xy$-plane is ${\cal D}$ given by
${\cal D} = \left\{ {\left( {x,y} \right)|0 \le x \le a,0 \le y \le b - \frac{b}{a}x} \right\}$
So,
${\cal W} = \left\{ {\left( {x,y,z} \right)|0 \le x \le a,0 \le y \le b - \frac{b}{a}x,0 \le z \le c - \frac{c}{a}x - \frac{c}{b}y} \right\}$
Evaluate the volume of ${\cal W}$:
$V = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} {\rm{d}}V = \mathop \smallint \limits_{x = 0}^a \mathop \smallint \limits_{y = 0}^{b - \frac{b}{a}x} \mathop \smallint \limits_{z = 0}^{c - \frac{c}{a}x - \frac{c}{b}y} {\rm{d}}z{\rm{d}}y{\rm{d}}x$
$ = \mathop \smallint \limits_{x = 0}^a \mathop \smallint \limits_{y = 0}^{b - \frac{b}{a}x} \left( {z|_0^{c - \frac{c}{a}x - \frac{c}{b}y}} \right){\rm{d}}y{\rm{d}}x$
$ = \mathop \smallint \limits_{x = 0}^a \mathop \smallint \limits_{y = 0}^{b - \frac{b}{a}x} \left( {c - \frac{c}{a}x - \frac{c}{b}y} \right){\rm{d}}y{\rm{d}}x$
$ = \mathop \smallint \limits_{x = 0}^a \left( {\left( {\left( {c - \frac{c}{a}x} \right)y - \frac{c}{{2b}}{y^2}} \right)|_0^{b - \frac{b}{a}x}} \right){\rm{d}}x$
$ = \mathop \smallint \limits_{x = 0}^a \left( {\left( {c - \frac{c}{a}x} \right)\left( {b - \frac{b}{a}x} \right) - \frac{c}{{2b}}{{\left( {b - \frac{b}{a}x} \right)}^2}} \right){\rm{d}}x$
$ = \mathop \smallint \limits_{x = 0}^a \left( {cb{{\left( {1 - \frac{1}{a}x} \right)}^2} - \frac{{cb}}{2}{{\left( {1 - \frac{1}{a}x} \right)}^2}} \right){\rm{d}}x$
$ = \frac{{cb}}{2}\mathop \smallint \limits_{x = 0}^a {\left( {1 - \frac{1}{a}x} \right)^2}{\rm{d}}x$
$ = \frac{{cb}}{2}\mathop \smallint \limits_{x = 0}^a \left( {1 - \frac{2}{a}x + \frac{1}{{{a^2}}}{x^2}} \right){\rm{d}}x$
$ = \frac{{cb}}{2}\left( {\left( {x - \frac{1}{a}{x^2} + \frac{1}{{3{a^2}}}{x^3}} \right)|_0^a} \right)$
$ = \frac{{cb}}{2}\left( {\frac{a}{3}} \right) = \frac{{abc}}{6}$
Evaluate the average of $z$-coordinate:
$\bar z = \frac{1}{V}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} z{\rm{d}}V = \frac{6}{{abc}}\mathop \smallint \limits_{x = 0}^a \mathop \smallint \limits_{y = 0}^{b - \frac{b}{a}x} \mathop \smallint \limits_{z = 0}^{c - \frac{c}{a}x - \frac{c}{b}y} z{\rm{d}}z{\rm{d}}y{\rm{d}}x$
$ = \frac{3}{{abc}}\mathop \smallint \limits_{x = 0}^a \mathop \smallint \limits_{y = 0}^{b - \frac{b}{a}x} \left( {{z^2}|_0^{c - \frac{c}{a}x - \frac{c}{b}y}} \right){\rm{d}}y{\rm{d}}x$
$ = \frac{3}{{abc}}\mathop \smallint \limits_{x = 0}^a \mathop \smallint \limits_{y = 0}^{b - \frac{b}{a}x} {\left( {c - \frac{c}{a}x - \frac{c}{b}y} \right)^2}{\rm{d}}y{\rm{d}}x$
$ = \frac{{3c}}{{ab}}\mathop \smallint \limits_{x = 0}^a \mathop \smallint \limits_{y = 0}^{b - \frac{b}{a}x} {\left( {1 - \frac{1}{a}x - \frac{1}{b}y} \right)^2}{\rm{d}}y{\rm{d}}x$
$ = \frac{{3c}}{{ab}}\mathop \smallint \limits_{x = 0}^a \mathop \smallint \limits_{y = 0}^{b - \frac{b}{a}x} \left( {{{\left( {1 - \frac{1}{a}x} \right)}^2} - \frac{2}{b}y\left( {1 - \frac{1}{a}x} \right) + \frac{{{y^2}}}{{{b^2}}}} \right){\rm{d}}y{\rm{d}}x$
$ = \frac{{3c}}{{ab}}\mathop \smallint \limits_{x = 0}^a \left( {\left( {{{\left( {1 - \frac{1}{a}x} \right)}^2}y - \frac{1}{b}\left( {1 - \frac{1}{a}x} \right){y^2} + \frac{{{y^3}}}{{3{b^2}}}} \right)|_0^{b - \frac{b}{a}x}} \right){\rm{d}}x$
$ = \frac{{3c}}{{ab}}\mathop \smallint \limits_{x = 0}^a \left( {{{\left( {1 - \frac{1}{a}x} \right)}^2}\left( {b - \frac{b}{a}x} \right) - \frac{1}{b}\left( {1 - \frac{1}{a}x} \right){{\left( {b - \frac{b}{a}x} \right)}^2} + \frac{1}{{3{b^2}}}{{\left( {b - \frac{b}{a}x} \right)}^3}} \right){\rm{d}}x$
$ = \frac{{3c}}{{ab}}\mathop \smallint \limits_{x = 0}^a \left( {b{{\left( {1 - \frac{1}{a}x} \right)}^3} - b{{\left( {1 - \frac{1}{a}x} \right)}^3} + \frac{b}{3}{{\left( {1 - \frac{1}{a}x} \right)}^3}} \right){\rm{d}}x$
$ = \frac{{3c}}{{ab}}\mathop \smallint \limits_{x = 0}^a \frac{b}{3}{\left( {1 - \frac{1}{a}x} \right)^3}){\rm{d}}x$
$ = \frac{c}{a}\mathop \smallint \limits_{x = 0}^a {\left( {1 - \frac{1}{a}x} \right)^3}{\rm{d}}x$
$ = \frac{c}{a}\mathop \smallint \limits_{x = 0}^a \left( {1 - \frac{{3x}}{a} + \frac{{3{x^2}}}{{{a^2}}} - \frac{{{x^3}}}{{{a^3}}}} \right){\rm{d}}x$
$ = \frac{c}{a}\left( {x - \frac{{3{x^2}}}{{2a}} + \frac{{{x^3}}}{{{a^2}}} - \frac{{{x^4}}}{{4{a^3}}}} \right)|_0^a$
$ = \frac{c}{a}\left( {a - \frac{{3a}}{2} + a - \frac{a}{4}} \right) = \frac{c}{4}$
So, $\bar z = \frac{c}{4}$.
Using the symmetry of the equation $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$, we conclude that $\bar x = \frac{a}{4}$ and $\bar y = \frac{b}{4}$.
Thus, the coordinates of the centroid: $\left( {\bar x,\bar y,\bar z} \right) = \left( {\frac{a}{4},\frac{b}{4},\frac{c}{4}} \right)$.
