 Three-colour clause gadget

Take auxiliary graph triangles $a,b,c$ and $d,t,e$, join $a$ to $d$, and add edges $bx,cy,ez$. When $t$ has colour $T$ and the inputs $x,y,z$ have colours in $\{T,F\}$, this graph admits an extension to a graph colouring with three colours exactly when at least one input is $T$. All-false inputs force $a=d=F$, contradicting their edge. Conversely, use $(a,b,c,d,e)=(T,F,B,F,B)$ if $x=T$, $(T,B,F,F,B)$ if $x=F,y=T$, and $(F,T,B,B,F)$ if $x=y=F,z=T$. The extension property is necessary for both directions of a polynomial-time many-one reduction.