As luksen and Marek point out, when the incoming particle is massless this reaction can't go, unless there's a third particle around to soak up the momentum. If we allow for the existence of this particle, then I don't think there's enough information to solve.
Let $p_{Xi},p_{Xf}$ be the initial and final 4-momenta of that particle. Then conservation of 4-momentum says
$$
p_\gamma+p_{Xi}=p_1+p_2+p_{Xf}
$$
Assume that the extra particle was initially at rest: $p_{Xi}=(M,0,0,0)$. ($M$ is the mass of $X$, and I'm setting $c=1$.) We don't know anything about any of the components of $p_\gamma$ or $p_{Xf}$, so there are eight unknowns to solve for. The above is 4 equations, so we need 4 additional equations. Unfortunately, we have only two,
$$
E_\gamma=|\vec p_\gamma|,
$$
$$
E_{Xf}^2=|\vec p_{Xf}|^2+M^2
$$
(Here $E$ means energy and $\vec p$ means 3-momentum in the frame we're working in, of course. These two equations are just $p_\mu p^\mu=m^2$, if you prefer.) So there's not enough information to solve.
The typical assumption is that the extra particle is much more massive than the others. In this case, the energy it absorbs is typically negligible, but the 3-momentum isn't. So if the 3-momentum is what you're interested in, I don't think you can get by with just pretending that particle isn't there.
This is a generic phase diagram :
I looked up the construction of a demonstration for classes cloud chamber using alcohol. For the air/alcohol gas line 3, constant pressure, change in temperature is where you want to work for your cloud chamber. Cooling it without condensation. The gradient of temperature in your chamber will define the thickness over the iced bottom where supersaturation can be maintained. So your choice of alcohol should depend on how slowly the phase diagram changes with temperature as to have a larger distance from the cold plate where the vapors will be supersaturated and tracks can form.
I do not think it is the gradient in the chamber that is decisive, but the one from the triple point to the vapor phase in line 3.The hot on top is to generate the vapour phase in the chamber. It will all depend on the phase diagram of your specific choice for vapor.
Best Answer
Basics
You need to
Getting a pair conversion event
Pair creation calls for the highest energy gamma you can get and as much mass in the chamber as you can arrange.
The odds of getting a pair-conversion event are graphed in figure 31.17 of the 2013 Review of Particle Physics chapter on the passage of radiation through matter, but it doesn't hit 10% in air until somewhere between 6 and 10 MeV. While this process can happen as soon as you get energies above $2m_e \approx 1.022 \,\mathrm{MeV}$, you need some energy left over to give the created particles some momentum. The mass attenuation length is graphed in figure 31.16 of the same reference, but for a few MeV gammas you are looking at distances around 20 cm.
Being able to tell that you got it
To be able to tell that you got a pair conversion event you need long enough tracks to convince yourself that you have a isolated "vee" and to tell that the tracks curve in opposite directions. That calls for a long enough propagation distance for the you to observer that the track is line-like and to see a non-trivial curvature. For arguments sake let's say that a one-tenth radian curve is good enough. That means getting tracks that are at least one-tenth their radius of curvature.
The radius of curvature of a particle in a magnetic field is given by $$ r = \frac{p}{qB} \,.$$ (Note that I don't write $p=mv$ because the pair may be at least moderately relativistic. Just stick to momentum.)
Engineering concerns
You need to pick a source of gamma rays, and you want it as energetic as possible. For really energetic gammas you need an accelerator based system, but this makes the project many times as big as it started. I'm going to assume you will use a radiological source, despite the low energy and unpredictable timing. Cobalt-60 would be enough if you have patience, but I'd suggest Thorium-232 if you can get it (you'll actually be taking advantage of the high energy gamma from the daughter isotope Thallium-208).
Using radiological sources means that you need a continuous data-acquisition systems of some kind--say digital video.
Finally you have to chose the strength of the magnetic field, and that depends on the expected momentum of your pair, which depends on the energy of your source.