= Solution
Use the two partials on each plane as obtained above. The four pairwise sums, in units of $a/6$, are
$$
\begin{array}{c|c|c}
\text{first-plane vector}+\text{second-plane vector}
&\text{resultant}&|\text{resultant}|^2\\\hline
[211]+[\bar211]&[022]&8\\
[211]+[\bar12\bar1]&[130]&10\\
[1\bar12]+[\bar211]&[\bar103]&10\\
[1\bar12]+[\bar12\bar1]&[011]&2
\end{array}
$$
The shortest resultant gives the largest line-energy reduction in the $Gb^2$ approximation:
$$
\boxed{\frac a6[1\bar12]+\frac a6[\bar12\bar1]
\longrightarrow\frac a6[011].}
$$
The initial summed squared magnitude is $a^2/3$ and the final value is $a^2/18$, a sixfold reduction in that part of the energy estimate.
The line of intersection of the two original planes is parallel to
$$
[\bar111]\times[111]=[0,2,-2]\parallel[01\bar1].
$$
Its dot product with the resultant <Burgers vector> is zero, so the junction is edge in character. A plane containing both line and <Burgers vector> has normal $[100]$, rather than an FCC close-packed $\{111\}$ glide normal. The resulting <stair-rod dislocation> is therefore sessile for ordinary glide and can form a <Lomer-Cottrell lock>. It obstructs further <dislocation> motion and promotes <work hardening>.
This energetic comparison assumes comparable line lengths and elastic prefactors. Geometry, line character and stacking-fault geometry affect a complete junction-energy calculation, but the small resultant identifies the favorable candidate requested by the vector model.
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