解:$(1)\triangle ABC$关于$y$轴对称的$\triangle A_{1}B_{1}C_{1},$
$A(-2,$$0)$关于$y$轴对称的点$A_{1}(2,$$0),$
$B(-1,$$0)$关于$y$轴对称的点$B_{1}(1,$$0),$
$C(-1,$$2)$关于$y$轴对称的点$C_{1}(1,$$2)$
$\triangle A_{1}B_{1}C_{1}$关于直线$l$对称的$\triangle A_{2}B_{2}C_{2},$
$A_{1}(2,$$0)$关于直线$l$对称的点$A_{2}(4,$$0),$
$B_{1}(1,$$0)$关于直线$l$对称的点$B_{2}(5,$$0),$
$C_{1}(1,$$2)$关于直线$l$对称的点$C_{2}(5,$$2)$
∴$\triangle A_{2}B_{2}C_{2}$三个顶点的坐标分别
是$A_{2}(4,$$0),$$B_{2}(5,$$0),$$C_{2}(5,$$2)$
$(2)$∵点$M$的坐标是$(3,$$0),$∴$OM = 3$
分类讨论如下:
如图①,如果$0<a\leqslant 3,$$P_{1}$在线段$OM$上
$PP_{2}=PP_{1}+P_{1}\ \mathrm {P}2=2OP_{1}+2P_{1}M $
$= 2(OP_{1}+P_{1}M)=2OM = 6$
如图②,如果$a>3,$点$P_{1}$在点$M$的右边
$PP_{2}=PP_{1}-P_{1}\ \mathrm {P}2=2OP_{1}-2\ \mathrm {P}1M $
$= 2(OP_{1}-P_{1}M)=2OM = 6$
综上,$PP_{2}$的长是$6$