The three-subgroup containment for normal subgroups

Put $P=x[[x^{-1},y],z]x^{-1}$, $Q=z[[z^{-1},x],y]z^{-1}$ and $R=y[[y^{-1},z],x]y^{-1}$. Direct substitution gives $P=yx{y}^{-1}zy{x}^{-1}{y}^{-1}x{z}^{-1}{x}^{-1}$, $Q=xz{x}^{-1}yx{z}^{-1}{x}^{-1}z{y}^{-1}{z}^{-1}$, and $R=zy{z}^{-1}xz{y}^{-1}{z}^{-1}y{x}^{-1}{y}^{-1}$. Cancelling adjacent inverse pairs in $PQ$ gives $yx{y}^{-1}zy{z}^{-1}{x}^{-1}z{y}^{-1}{z}^{-1}=R^{-1}$. Thus $PQR=1$.

For normal $U,V$, $[U,V]$ is normal: conjugation sends its generator $[u,v]$ to $[gu{g}^{-1},gv{g}^{-1}]$. Therefore $N=[[B,C],A][[C,A],B]$ is a normal subgroup (a product of two normal subgroups is a subgroup since its factors can be interchanged). Work in $G\mathrm{/}N$. Substituting $x\in A,y\in B,z\in C$ into step 1.1 makes $Q=R=1$, so $[[x^{-1},y],z]=1$. Replacing $x$ by its inverse shows that every $[a,b]$ commutes with every $c\in C$ in this quotient.

If $u$ and $v$ commute with every $c\in C$ in the quotient, the product identity gives $[uv,c]=u[v,c]u^{-1}[u,c]=1$; and $[u^{-1},c]=1$ follows from $1=[u{u}^{-1},c]$. Consequently every finite product of the generators $[a,b]$ and their inverses centralizes $C$. Hence $[[A,B],C]$ has trivial image in $G\mathrm{/}N$, which is precisely the claimed containment. This includes any of $A,B,C$ equal to $1$.