Deriving The Triple Point Temperature and Pressure as the following expression
On a phase diagram there are three 1-dimensional curves, the melting curve, the liquid vaporization curve, and the sublimation curve.  

It is sufficient to consider just two of the three curve, and determine the point at which they intersect.  

Given the limited information availible to you, (no information about how volume changes upon going from the solid to liquid phase), it is wise to choose the liquid vaporization curve and sublimation curve, rather than the melting curve, as $\Delta V$ can be approximated as the volume of gas, the volume of solid or liquid being approximately zero compare to the much larger volume of gas.

From the Clapeyron equation in the question, you need to change variables from  $\Delta S$ which you don't know to $\Delta H$ which you do know. 

When two phases are in equilibrium, such as along any of the three curves discussed above, $\Delta G$ = 0.

$G = H-TS$

$0 = dG = dH-TdS -SdT$

Considering that $T$ is constant during a phase change:

$\Delta H = T \Delta S$

So substituting:

$\frac{dP}{dT}= \frac{ΔH}{TΔv}$

$\frac{dP}{dT}= \frac{ΔH_{vap}}{TΔv}$ and $\frac{dP}{dT}= \frac{ΔH_{sub}}{TΔv}$

Then, as mentioned above the approximation $\Delta V = V_{gas}$ can be made and the ideal gas law (or other equation of state) can be used to express $V$ as a function of $T$ and $P$.  Then (after intergrating) you have two equations with two unknowns, $T$ an $P$, the temperature and pressure at the triple point.