Compute the arithmetic genus of the union of the three coordinate axes Z(x_1 x_2, x_1 x_3, x_2 x_3) subseteq P^3.

Step 1
There is a nice general method for calculating the arithmetic genus of a reducible curve like this where a bunch of curves are being glued transversely. It uses the (equivalent) description of the arithmetic genus of a curve X over a field as the dimension of the cohomology group $H^1({\mathcal{O}}_X)$.
So suppose $C_1$ and $C_2$ are curves over a field. Let X denote the curve we get by glueing a point $p_1\in <!--\; \in \; -->C_1$ transversely to a point $p_2\in <!--\; \in \; -->C_2$. Call the image of the glueing points $p\in <!--\; \in \; -->X$. Then we have an exact sequence of sheaves on X of the form
$$0\to <!--\; \to \; -->O_X\to <!--\; \to \; -->{\mathcal{O}}_{C_1}\oplus <!--\; \oplus \; -->{\mathcal{O}}_{C_2}\to <!--\; \to \; -->k_p\to <!--\; \to \; -->0$$
where $k_p$ is the skyscraper sheaf supported at p. (The first map is pullback of functions; the second is zero away from p, and $(f_1,f_2)\mapsto <!--\; \mapsto \; -->f_1(p)-<!--\; -\; -->f_2(p)$ on open sets containing p.)
Step 2
Now take cohomology of this sequence: you get
$$0\to <!--\; \to \; -->H^0(O_X)\to <!--\; \to \; -->H^0({\mathcal{O}}_{C_1})\oplus <!--\; \oplus \; -->H^0({\mathcal{O}}_{C_2})\to <!--\; \to \; -->H^0(k_p)\to <!--\; \to \; -->H^1(O_X)\to <!--\; \to \; -->H^1({\mathcal{O}}_{C_1})\oplus <!--\; \oplus \; -->H^1({\mathcal{O}}_{C_2})\to <!--\; \to \; -->0$$
where the final zero comes from the fact that is supported at a single point.
Now, I leave it to you to check that the first three nonzero terms give a short exact sequence. That implies that
$$H^1(O_X)=H^1({\mathcal{O}}_{C_1})\oplus <!--\; \oplus \; -->H^1({\mathcal{O}}_{C_2})$$
In other words, arithmetic genus is additive when we glue curves transversely at a point! Notice that we didn't require anything to be smooth or irreducible, so you can apply this in your situation in two steps by glueing two of the axes together, then glueing on the third.
Exercise: Modify the above argument to show that if Y is a curve obtained by transverse glueing of two points on the same connected curve X, then
$$\dim <!--\; \; -->H^1({\mathcal{O}}_Y)=\dim <!--\; \; -->H^1({\mathcal{O}}_X)+1$$
Use this to figure out what happens when you glue any number of curves transversely at any number of points!

Step 1
Here is an elementary, easy, self-contained proof which, as a friend reminded me, results from a discussion we had a few months ago. I could kick myself for forgetting about it yesterday ...
Consider two subschemes .
$X_1,X_2\subset <!--\; \subset \; -->{\mathbb{P}}_k^n=\mathrm{Proj}<!--\; \; -->(S)(S=k[T_0,\cdots <!--\; \cdots \; -->,T_n])$
From the exact sequence
$$0\to <!--\; \to \; -->S/I(X_1)\cap <!--\; \cap \; -->I(X_2)\to <!--\; \to \; -->S/I(X_1)\oplus <!--\; \oplus \; -->S/I(X_2)\to <!--\; \to \; -->S/(I(X_1)+I(X_2))\to <!--\; \to \; -->0$$
we get the relation between Hilbert polynomials
$$H_{X_1}(t)+H_{X_2}(t)=H_{X_1\cup <!--\; \cup \; -->X_2}(t)+H_{X_1\cap <!--\; \cap \; -->X_2}(t)(\mathrm{★<!--\; ★\; -->})$$
Applying this to
$$X_1=V(x_3,x_1{x}_2),X_2=V(x_1,x_2),X_1\cap <!--\; \cap \; -->X_2=V(x_1,x_2,x_3)=\{[1:0:0:0]\},X=X_1\cup <!--\; \cup \; -->X_2$$
we get, remembering that $X_1$ is just a reducible plane conic so that $H_{X_1}(t)=2t+1$, that $X_2$ is a line so that $H_{X_2}(t)=t+1$, and that $X_1\cap <!--\; \cap \; -->X_2$ is a reduced point with Hilbert polynomial the constant 1:
$$(2t+1)+(t+1)=H_X(t)+(1)(\mathrm{★<!--\; ★\; -->}\mathrm{★<!--\; ★\; -->})$$
Hence $H_X(t)=3t+1$ and thus $p_a(X)=0$.
Step 2
WARNING
One might wonder why (fortunately!) the above calculation doesn't work when $X_2^{\prime }$ is a line in the same plane as $X_1$, going through the singularity P of $X_1$.
The answer is that in this case the intersection $X_1\cap <!--\; \cap \; -->X_2^{\prime }$ is no longer a simple point but the double point at P embedded in $X_2^{\prime }$.
In that case the analogous calculation yields for $X_0=X_1\cup <!--\; \cup \; -->X_2^{\prime }$ the Hilbert polynomial $H_{X_0}(t)=3t$, as already mentioned in my other answer.
One must very carefully note that the formula $(\mathrm{★<!--\; ★\; -->})$ is correct only if one interprets union and intersection in the scheme-theoretic sense: never ever forget about nilpotents!