= Solution
The FCC parity rule forbids $(100)$ and $(110)$ and allows $(200)$ and $(111)$. To follow each same plane family through the deformation, use fractional coordinates $(x,y,z)$ in the compressed old cell and $(X,Y,Z)$ in the new cube. The preceding vectors give
$$
x=(X-Y)/2,\qquad y=(X+Y)/2,\qquad z=Z.
$$
Thus $hx+ky+lz=m$ becomes $HX+KY+LZ=m$, with
$$
\boxed{(H,K,L)=\left(\frac{h+k}{2},\frac{k-h}{2},l\right).}
$$
Crucially, this transformation retains the plane spacing and diffraction order, not just a normal direction:
|| Old FCC reflection
|| New-cell reciprocal coordinates
|| Status
| $(100)$
| $(\tfrac12,-\tfrac12,0)$
| Not a new reciprocal-lattice vector; absent
| $(110)$
| $(1,0,0)$
| Odd index sum; absent
| $(200)$
| $(1,-1,0)$
| Even index sum; allowed
| $(111)$
| $(1,0,1)$
| Even index sum; allowed
The two physically allowed reflections remain allowed, and the two absent ones remain absent. Their angles and spacings change under strain; their allowedness as tracked reciprocal vectors does not.
The half-integer entry is deliberate. It cannot be relabeled as the ordinary BCC $(1\bar10)$ reflection while claiming to track the same diffraction order. Multiplying its indices by two doubles the reciprocal vector and halves the spacing; that new allowed reflection is the image of old FCC $(200)$, not old $(100)$. If only an orientation label is wanted, the old $(100)$ and $(200)$ planes both have the new $(1\bar10)$ orientation, but are distinct orders. This is the distinction between <diffraction indexing under a cell change> and merely reducing a plane normal to the smallest integer ratio.
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