= Solution
The event rate counts unordered pairs of reacting <nuclei>. For unlike species there are $n_in_j$ pairs per unit-volume normalization; for identical species there are approximately $n_i^2/2$ pairs, since counting ordered pairs counts each collision twice. The <Kronecker delta> therefore supplies exactly the required divisor $1+\delta_{ij}$. It is the event rate, not the destruction rate of identical nuclei: one identical-pair reaction destroys two nuclei.
The <temperature> dependence combines the high-energy tail of the <Maxwell-Boltzmann distribution> with <quantum tunnelling> through the <Coulomb barrier>. The principal exponential factor in the energy integral is $\exp[-E/(k_BT)-b/\sqrt E]$. Its stationary point, the <Gamow peak>, has $E_0\propto T^{2/3}$ and exponent $\eta=3E_0/(k_BT)\propto T^{-1/3}$. Integrating around this peak gives a prefactor proportional to $T^{-2/3}$, or $\eta^2$, when the nuclear factor varies slowly. Thus the supplied <thermonuclear reaction rate> is locally proportional to $\eta^2e^{-\eta}$. The slow weak interaction in the proton-proton reaction also makes its overall coefficient very small; its local thermal slope follows the supplied formula.
At fixed <number densities>, define the <local temperature exponent of a thermonuclear reaction> by a logarithmic derivative. Since $d\eta/d\log T=-\eta/3$,
$$
\frac{d\log\lambda_{ij}}{d\log T}=(2/\eta-1)(-\eta/3)=\frac{\eta-2}{3}.
$$
For proton-proton reactions, $A=1/2$ and $Z_1=1$. At $T_6=15$, $\eta_{11}=13.67$ and $\alpha=3.89$. For two <helium-3> nuclei, $A=3/2$ and $Z_3^2Z_3^2=16$, giving $\eta_{33}=49.68$ and $\beta=15.89$. For <helium-3> with <helium-4>, $A=12/7$ and the same charge product gives $\eta_{34}=51.95$ and $\gamma=16.65$. Hence
$$
\boxed{\alpha\simeq4,\qquad\beta\simeq16,\qquad\gamma\simeq16,\quad\gamma>\beta}.
$$
These are local powers near the specified temperature, not constant exponents over an arbitrarily large range. The greater reduced nuclear mass in the $34$ reaction increases its tunnelling exponent and its thermal sensitivity.
For the <effective proton-proton reaction network>, denote the event rates by
$$
R_{11}=\tfrac12\lambda_{11}n_1^2,\quad R_{21}=\lambda_{21}n_2n_1,\quad R_{33}=\tfrac12\lambda_{33}n_3^2,\quad R_{34}=\lambda_{34}n_3n_4,\quad R_{17}=\lambda_{17}n_1n_7.
$$
The short-lived beryllium intermediate can be eliminated, so production through $R_{34}$ feeds $n_7$, representing predominantly lithium-7. This rapid conversion is <electron capture>. Count the nuclei consumed and produced in each reaction: $11$ consumes two protons and makes one <deuterium>; $21$ consumes a proton and a <deuterium> and makes one <helium-3>; $33$ consumes two <helium-3> and makes one <helium-4> and two protons; $34$ consumes one of each helium isotope; $17$ consumes a proton and lithium-7 and makes two <helium-4>. It follows that
$$
\begin{aligned}
\dot n_1&=-\lambda_{11}n_1^2-\lambda_{21}n_2n_1+\lambda_{33}n_3^2-\lambda_{17}n_1n_7,\\
\dot n_2&=\tfrac12\lambda_{11}n_1^2-\lambda_{21}n_2n_1,\\
\dot n_3&=\lambda_{21}n_2n_1-\lambda_{33}n_3^2-\lambda_{34}n_3n_4,\\
\dot n_4&=\tfrac12\lambda_{33}n_3^2-\lambda_{34}n_3n_4+2\lambda_{17}n_1n_7.
\end{aligned}
$$
For closure, $\dot n_7=\lambda_{34}n_3n_4-\lambda_{17}n_1n_7$. These nuclear terms conserve $n_1+2n_2+3n_3+4n_4+7n_7$, providing an independent stoichiometric check. They describe a fixed local volume; compression or mixing would add transport terms.
The roughly one-second destruction time of <deuterium> is negligible compared with the proton-burning and <helium-3> evolution times. Its abundance rapidly adjusts to the slowly evolving production rate. Set $\dot n_2\simeq0$, giving the <deuterium quasi-equilibrium in proton-proton burning>
$$
\lambda_{21}n_2n_1\simeq\tfrac12\lambda_{11}n_1^2,\qquad n_2\simeq\frac{\lambda_{11}n_1}{2\lambda_{21}}.
$$
Substitution, without setting $\dot n_1=0$, gives
$$
\boxed{\begin{aligned}
\dot n_1&\simeq-\tfrac32\lambda_{11}n_1^2+\lambda_{33}n_3^2-\lambda_{17}n_1n_7,\\
\dot n_3&\simeq\tfrac12\lambda_{11}n_1^2-\lambda_{33}n_3^2-\lambda_{34}n_3n_4.
\end{aligned}}
$$
The <helium-3 relaxation time> of about $6\times10^5$ years is likewise short compared with the central proton-burning time and the age of the <Sun>. On this relaxation time the proton and <helium-4> abundances, and the background temperature, can be treated as fixed. Put $A=\lambda_{33}$, $B=\lambda_{34}n_4$, $C=\lambda_{11}n_1^2/2$. Then $\dot n_3=C-An_3^2-Bn_3$, and its nonnegative equilibrium is
$$
\boxed{n_{3e}=-\frac{\lambda_{34}n_4}{2\lambda_{33}}+\sqrt{\left(\frac{\lambda_{34}n_4}{2\lambda_{33}}\right)^2+\frac{\lambda_{11}n_1^2}{2\lambda_{33}}}}.
$$
The other root is negative and unphysical. Write $n_3=n_{3e}+x$ and subtract the equilibrium equation. Exactly, for a fixed background,
$$
\dot x=-(2\lambda_{33}n_{3e}+\lambda_{34}n_4)x-\lambda_{33}x^2.
$$
Dropping the quadratic displacement gives
$$
\boxed{\dot x=-x/\tau,\qquad\tau=(2\lambda_{33}n_{3e}+\lambda_{34}n_4)^{-1}}.
$$
Thus small displacements decay exponentially. A slowly moving equilibrium introduces an additional $-\dot n_{3e}$ term, negligible to leading order when $\tau$ is short compared with background evolution.
To estimate the <temperature for helium-3 freeze-out>, use the cooler pp-I-dominated region and neglect $B$ compared with $2An_{3e}$. Then $n_{3e}\simeq n_1\sqrt{\lambda_{11}/(2\lambda_{33})}$ and $\tau^{-1}\simeq n_1\sqrt{2\lambda_{11}\lambda_{33}}$. With the local powers just obtained and fixed proton <number density>, $n_{3e}\propto T^{-6}$ and $\tau\propto T^{-10}$. Adopting $t_\odot\simeq4.6\times10^9\,\mathrm{yr}$ gives
$$
T\simeq15\times10^6\left(\frac{6\times10^5}{4.6\times10^9}\right)^{1/10}\!\mathrm K\simeq\boxed{6.1\times10^6\,\mathrm K}.
$$
This is an order-of-magnitude extrapolation of local exponents. Keeping the supplied exponential factors gives the <helium-3 freeze-out estimate with Gamow factors>
$$
\frac{\tau(T)}{\tau(T_c)}=\left(\frac{T}{T_c}\right)^{2/3}\exp\!\left[\frac{\eta_{11,c}+\eta_{33,c}}2\left(\left(\frac{T_c}{T}\right)^{1/3}-1\right)\right],
$$
which yields about $6.8\times10^6\,\mathrm K$ with the same normalization and fixed-density approximation. Thus the robust estimate is several million kelvin. The supplied data do not fix the local density, composition or pp-II fraction away from the centre, so they cannot determine an exact transition temperature or radius. In particular, using $\tau\propto T^{-16}$ alone would omit the temperature dependence of the equilibrium <helium-3> abundance.
The <solar hydrogen and helium-3 abundance profiles> follow from burning, finite equilibration time and <convection>. Central <hydrogen> has been consumed, so $X_1$ is lowest in the centre and increases toward its much less processed outer value. In the hot centre, rapid destruction keeps <helium-3> small. Moving outward initially raises its equilibrium abundance because $\lambda_{11}/\lambda_{33}$ rises strongly as temperature falls. Further out, equilibration is no longer reached within the solar age, and still farther out even production is slow. Hence $X_3$ has an off-centre maximum and then declines toward its small outer abundance. The outer convective envelope mixes each abundance toward a constant value.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2003/iii/paper-63-solar-abundances.png]
{title=Present solar hydrogen depletion and the off-centre helium-3 abundance maximum}
The sketch shows these qualitative shapes; the helium-3 ordinate is scaled to its own maximum. The illustrated amplitudes, peak radius and envelope boundary are schematic, not a numerical solar-model solution.
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