A very straightforward way is to use a triple integral $$\begin{array}{r}P(X<Y<Z)={\int }_0^{\mathrm{\infty }}{\int }_0^z{\int }_0^y\lambda e^{-\lambda x}\mu e^{-\mu y}\nu e^{-\nu z}dxdydz=\frac{\lambda \mu }{(\lambda +\mu +\nu )(\mu +\nu )}\end{array}$$ $P(X<Y<Z)$ can be broken down as $P(X<min(Y,Z))P(Y<Z)$.

Consider just $P(Y<Z)$ $$\begin{array}{r}P(Y<Z)={\int }_0^{\mathrm{\infty }}{\int }_0^z\mu e^{-\mu y}\nu e^{-\nu z}dydz=\frac{\mu }{\mu +\nu }\end{array}$$ Thus, when two exponential processes are considered, probaility of arrival of 1st before 2nd is simply the percentage ratio of parameters. Thus, $$\begin{array}{rl}P(X<min(Y,Z))& =\frac{\lambda }{\lambda +(\mu +\nu )}Y\text{ and }Z\text{ can be combined as a single process}\\ P(Y<Z)& =\frac{\mu }{\mu +\nu }\\ P(X<Y<Z)& =P(X<min(Y,Z))P(Y<Z)\\ & =\frac{\lambda \mu }{(\lambda +\mu +\nu )(\mu +\nu )}\end{array}$$

where we have also used the property that minimum of exponentially distributed random variables is itself exponentially distributed.