The equations of the sides $AB$,$AC$,and $BC$ of a $\triangle ABC$ are respectively $x-3y=0$,$3x-y=0$,and $x+y+4=0$. If $P$ and $Q$ are points on the line $3x-y+k=0$ passing through $B$ such that $PB:BQ=1:1$,then $k=$
- A$8$
- B$12$
- C$-8$
- D$-12$
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