pman

解:如图:∵S△PBC=

1
2
PM?BC,S△ABC=
1
2
AN?BC,

S△PBC
S△ABC
=
PM
AN
=
PD
AD
=
x
x+6

同理:

S△PAC
S△ABC
=
y
y+6
S△PAB
S△ABC
=
z
z+6

∵S△ABC=S△PBC+S△PCA+S△PAB,

x
x+6
+
y
y+6
+
z
z+6
=1.

即1-

6
x+6
+1-
6
y+6
+1-
6
z+6
=1,

3
x+6
+
3
y+6
+
3
z+6
=1,

∴3(yz+zx+xy)+36(x+y+z)+324

=xyz+6(xy+yz+zx)+36(x+y+z)+216,

∴xy+yz+zx=28.

∴xyz=108-3(xy+yz+zx)=24.

答:xyz的大小为:24.