高二数学...急急急...

距离d=|-2(2+k)-2(1+k)-2(3+2k)|/√[(2+k)^2+(1+k)^2]
=|-4-2k-2-2k-6-4k|/√(4+4k+k^2+1+2k+k^2)
=|8k+12|/√(2k^2+6k+5)

令m=d^2=(64k^2+192k+144)/(2k^2+6k+5)
64k^2+192k+144=2mk^2+6mk+5m
(64-2m)k^2+(192-6m)k+144-5m=0
这个关于k的方程有解
所以判别式=(192-6m)^2-4(64-2m)(144-5m)>=0
36864-2304m+36m^2-36864+2432m-40m^2>=0
m^2-32m<=0
0<=m<=32
0<=d^2<=32
0<=d<=4根号2

所以无论k取什么值，P到直线距离不大于 4根号2