Y1= m1x-3m1+2 when x=a, y1= am1-3m1+2 = 2+(a-3) m/kloc.

y2=m2x-3m2+2，y2 = am2-3m 1+2 = 2+(a-3)m 1

(1) when a=3, y 1=y2=2.

(2) when a >; At 3 o'clock, (a-3) m 1

Then p = y 1 = 2+(a-3) m 1

(3) when a

Then p = y2 = 2+(a-3) m 1

Therefore, the maximum value of p is 2.