= Solution
The original PDF contains three algebraic misprints in this question. The second internal <four-momentum> must consistently be $r'_k$, since it is defined by $r'_k=q_k-p'=r_i-q_j$. Its <Dirac propagator> denominator is $(r'_k)^2-m^2$, including after contracting the first photon; the printed plus sign is incorrect. Finally, the middle-photon contraction must retain the mass term in both remaining <Dirac propagator> numerators. The following calculation keeps the electron mass and proves the corrected identities.
Write $r=p-q_i$, $R=q_k-p'=r-q_j$ and introduce
$$
D(l)=\not l-m,\qquad F(l)=D(l)^{-1}=\frac{\not l+m}{l^2-m^2}.
$$
Here $F$ is the matrix part of the <Dirac propagator>, with the overall factors of $i$ kept in the common amplitude convention. The <Clifford algebra> proves $D(l)F(l)=F(l)D(l)=I$. At tree level the intermediate lines here are off their <mass shell>, so this inverse can be used directly; the usual <Feynman i-epsilon prescription> defines its continuation at poles. With $C=(ie)^3$, the consistently routed tensor is
$$
T_{ijk}^{\mu\nu\sigma}=C\bar v(p')\gamma^\sigma F(R)\gamma^\nu F(r)\gamma^\mu u(p).
$$
The external <Dirac equations> are $(\not p-m)u(p)=0$ and $\bar v(p')(\not p'+m)=0$.
For the photon at the electron end, $\not q_i=\not p-\not r$ and hence $\not q_i u=-D(r)u$. Multiplication by $F(r)$ removes that internal line, giving
$$
\boxed{T_{ijk}^{\mu\nu\sigma}q_{i\mu}=-C\bar v(p')\gamma^\sigma F(R)\gamma^\nu u(p).}
$$
For the middle photon, $\not q_j=D(r)-D(R)$. Associativity, without any commuting of <gamma matrices>, gives
$$
F(R)\not q_j F(r)=F(R)D(r)F(r)-F(R)D(R)F(r)=F(R)-F(r).
$$
Therefore
$$
\boxed{T_{ijk}^{\mu\nu\sigma}q_{j\nu}=C\bar v(p')\gamma^\sigma\left[\frac{\not R+m}{R^2-m^2}-\frac{\not r+m}{r^2-m^2}\right]\gamma^\mu u(p).}
$$
Omitting the two mass numerators changes this expression by
$$
Cm\left[\frac1{R^2-m^2}-\frac1{r^2-m^2}\right]\bar v(p')\gamma^\sigma\gamma^\mu u(p),
$$
which need not vanish. For a concrete counterexample take $m=1$, $p=(2,0,0,\sqrt3)$, $p'= (2,0,0,-\sqrt3)$ and outgoing <photon> momenta $q_1=(1,1,0,0)$, $q_2=(6/5,2/5,4\sqrt2/5,0)$, $q_3=(9/5,-7/5,-4\sqrt2/5,0)$. All external momenta are <on shell> and conserve <four-momentum>. Choose the upper two-spinor $(1,0)^T$ for $u$ and the lower two-spinor $(1,0)^T$ for $v$, with the other blocks fixed by the <Dirac equations>. For $(i,j,k)=(1,2,3)$ and $(\mu,\sigma)=(0,3)$, the correct middle contraction divided by $C$ is $1/3$, whereas the expression with the mass numerators omitted is $5/27$. Thus the printed middle identity is not valid for a massive <electron>.
At the positron end, $q_k=R+p'$. The adjoint <Dirac equation> gives $\bar v\not q_k=\bar v(\not R-m)=\bar vD(R)$, so the remaining requested contraction is
$$
\boxed{T_{ijk}^{\mu\nu\sigma}q_{k\sigma}=C\bar v(p')\gamma^\nu F(r)\gamma^\mu u(p).}
$$
The opposite signs at the two ends are precisely what makes the <Ward identity> work.
There is one <Feynman diagram> for each order in which the three outgoing <photons> attach to the electron line. Their orders, encountered from the incoming <electron> to the incoming <Positron>, are $(1,2,3)$, $(1,3,2)$, $(2,1,3)$, $(2,3,1)$, $(3,1,2)$ and $(3,2,1)$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2002/iii/paper-62-three-photon-diagrams.png]
{title=The six tree diagrams for electron-positron annihilation into three photons, ordered along fermion flow}
The straight-line arrows indicate <fermion flow>; the incoming <Positron> momentum is opposite to its line's <fermion flow>. Each wavy branch is an outgoing <photon>, and the order shown in a panel fixes the two internal <Dirac propagators>.
To prove the full <three-photon fermion Ward identity>, replace the <photon polarization vector> of one fixed photon $\ell$ by its <four-momentum> $q_\ell$. Fix the order $(a,b)$ of the other two <photons> and write $\Gamma_a=\not\epsilon_a$, $\Gamma_b=\not\epsilon_b$. There are exactly three positions for photon $\ell$ relative to that order. By the three contractions just derived, their contributions, divided by $C$, are
$$
\begin{aligned}
(\ell,a,b):\quad&-\bar v\Gamma_b F(q_b-p')\Gamma_a u,\\
(a,\ell,b):\quad&\bar v\Gamma_b[F(q_b-p')-F(p-q_a)]\Gamma_a u,\\
(a,b,\ell):\quad&+\bar v\Gamma_b F(p-q_a)\Gamma_a u.
\end{aligned}
$$
<Four-momentum conservation> $p+p'=q_\ell+q_a+q_b$ identifies the shared internal momenta in these expressions. The three terms cancel exactly. Repeating for the other order $(b,a)$ covers all six <Feynman diagrams>, and proves
$$
\boxed{\mathcal A(\epsilon_\ell\to q_\ell)=0\quad\text{for each }\ell=1,2,3.}
$$
Individual <Feynman diagrams> generally fail this test; their complete sum is essential. By linearity in each <photon polarization vector>, the result also proves $\mathcal A(\epsilon_\ell+cq_\ell)=\mathcal A(\epsilon_\ell)$. \b[The amplitude is gauge invariant:] a pure-gauge external <photon> polarization decouples, and the physical <scattering amplitude> depends only on the two transverse <photon> polarizations. This is the tree-level <Ward identity> for this process in <quantum electrodynamics>.
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