若A,B,C能构成三角形,
则AB,AC,BC相互不平行
AB=6I-3J-3I+4J=3I+J
AC=-3I+4J+(5-M)I-(4+M)J=(2-M)I-MJ
BC=-6I+3J+(5-M)I-(4+M)J=(-1-M)I+(-1-M)J
即(2-m)/3≠(-m)/1 解得M≠- 1
(-1-M)/3≠(-1-M)/1 解得M≠- 1
(-1-M)/(2-m)≠(-1-m)/(-m）解得M≠- 1
综上：实数M应满足M≠－1
2.AB⊥AC
则3(2－M)-M=0
M=3/2
AB⊥BC
则3(-1-M)-1-M=0
M=- 1(舍去)
AC⊥BC
则(2-M)(-1-M) +(-M)(- 1-M)=0
(2-2M)(-1-M)=0
M=1 ,M=- 1(舍去)
M=3/2或者1