Here is how to solve it without much calculus. Remember that for exponential variable $X$ with parameter $\lambda$, $\mathbb{P}(X > x) = \exp(-\lambda x)$ for $x \ge 0$.
We will make use of the following result, that relies on memoryless property of exponential distribution, i.e. $\mathbb{E}(f(X-a) ; X > a) = \mathbb{E}(f(X))$: $ \begin{eqnarray} \mathbb{E}( \mathbf{1}_{X > a} \exp(-\mu X)) &=& \mathbb{E}(\exp(-\mu X); X > a) \mathbb{P}(X > a) \\ &=& \exp(-\mu a) \cdot \mathbb{E}(\exp(-\mu (X-a)); X > a) \cdot \exp(-\lambda a) \\ &=& \exp(-(\mu+\lambda) a) \cdot \mathbb{E}(\exp(-\mu X) ) \\ &=& \exp(-(\mu+\lambda) a) \cdot \frac{\lambda}{\lambda+\mu} \end{eqnarray} $
Now, use conditioning: $ \begin{eqnarray} \mathbb{P}(C > B > A) &=& \mathbb{E}_A( \mathbb{E}_B( \mathbf{1}_{B>a} \mathbb{P}(C > b ; B=b, A=a) ; A=a)) \\ &=& \mathbb{E}_A( \mathbb{E}_B( \mathbf{1}_{B>a} \exp(-\lambda_c b) ; A=a)) \\ &=& \mathbb{E}_A( \frac{\lambda_b}{\lambda_b+\lambda_c} \exp(-a(\lambda_b+\lambda_c) ) \\ &=& \frac{\lambda_b}{\lambda_b+\lambda_c} \cdot \frac{\lambda_a}{\lambda_a + \lambda_b+\lambda_c} \end{eqnarray} $
With $\lambda_a=1$, $\lambda_b = 2$, $\lambda_c=3$, the answer comes to $\frac{1}{15}$.