若x-3=y-2=z-1,求x的平方+y的平方++z的平方-xy-yz-zx的值

x-3=y-2x-y=1
y-2=z-1y-z=1
x-z=2
x²+y²+z²-xy-yz-zx
=(1/2)(x²-2xy+y²+y²-2yz+z²+z²-2zx+x²)
=(1/2)[(x-y)²+(y-z)²+(x-z)²]
=(1/2)(1²+1²+2²)
=3