解：?$(1)M(12，$??$0)，$??$P(6，$??$6)$?
?$ (2)$?設(shè)二次函數(shù)表達(dá)式為?$y=a(x-6)^2+6$?
∵函數(shù)?$y=a(x-6)^2+6$?的圖像經(jīng)過點(diǎn)?$(0，$??$0)$?
∴?$0=a(0-6)^2+6，$?即?$a=- \frac {1}{6}$?
∴拋物線相應(yīng)的函數(shù)表達(dá)式為?$y=-\frac {1}{6} (x-6)^2+6，$?
即?$y=-\frac 16x^2+2x $?
?$(3)$?設(shè)?$A(m，$??$0)，$?則?$B(12-m，$??$0)，$??$C(12-m，$??$- \frac {1}{6}\ \mathrm {m^2}+2\ \mathrm {m})，$
??$D(m，$??$- \frac {1}{6}\ \mathrm {m^2} +2\ \mathrm {m})$?
∴“支撐架”總長?$AD+DC+CB= (- \frac {1}{6}\ \mathrm {m^2}+2m)+(12-2m) +(- \frac {1}{6}\ \mathrm {m^2}+2m )$?
? $=- \frac {1}{3}\ \mathrm {m^2}+2m+12=- \frac {1}{3} (m-3)^2+15$?
∵此二次函數(shù)的圖像開口向下
∴當(dāng)?$m=3\ \mathrm {m} $?時，?$AD+DC+CB$?有最大值為?$15\ \mathrm {m}$?
