Solution:

Put $f(x) = x(x-2)(3x-7) - 2 = 3x^{3} - 13x^{2} + 14x - 2$. Then $f(0) = -2$, $f(1) = 2$, so there is a root between $0$ and $1$. $f(2) = -2$, so there is another root between $1$ and $2$. $f(3) = 4$, so the third root is between $2$ and $3$. $f(x) = 0$ has three roots, so they are all real and positive.

We have $\tan(a + b + c) = \dfrac{\tan a + \tan b + \tan c - \tan a \tan b \tan c}{1 - (\tan a \tan b + \tan b \tan c + \tan c \tan a)}$. So putting $a = \tan^{-1} r$, $b = \tan^{-1} s$, $c = \tan^{-1} t$, we have

$$\tan(a + b + c) = \frac{(r + s + t) - rst}{1 - (rs + st + tr)} = \frac{13/3 - 2/3}{1 - 14/3} = \frac{11/3}{-11/3} = -1.$$

So $a + b + c = -\pi/4 + k\pi$. But we know that each of $r, s, t$ is real and positive, so $a + b + c$ lies in the range $0$ to $3\pi/2$. Hence $a + b + c = 3\pi/4$.