= Solution
Initially the two inner orbital normals coincide, so their combined <angular momentum> is $L_{12}=L_1+L_2$ along one direction. Use the outer orbital normal as the vertical axis. The inner pair's vector then has perpendicular component $L_{12}\sin\Delta I$ and parallel component $L_{12}\cos\Delta I$. Adding the outer <angular momentum> gives
$$
L_{\perp}=L_{12}\sin\Delta I,\qquad
L_{\parallel}=L_3+L_{12}\cos\Delta I.
$$
The <invariable plane> is perpendicular to this total vector. Thus the outer orbit's <orbital inclination> to that plane equals the angle between its own normal and the total normal. The <inclination to the invariable plane of two orbital planes> is
$$
\boxed{I_3=\arctan\left(\frac{\sin\Delta I}{A+\cos\Delta I}\right),\qquad A=\frac{L_3}{L_1+L_2}.}
$$
This uses the small, prograde inclinations for which the denominator is positive; in general use the corresponding two-argument arctangent to choose the quadrant. If $A\gg1$, $I_3\simeq\Delta I/(A+1)$ for small $\Delta I$: the distant planet carries most of the <angular momentum> and lies close to the <invariable plane>.
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