已知x+y=3,z-y=5,xy-yz+xz=7,那么x的平方+y的平方+z的平凡等于多少?

x+y=3,z-y=5,x+z=8(x+y)^2+(z-y)^2+(x+z)^2-2(xy-yz+xz)=x^2+y^2+2xy+z^2-2zy+y^2+x^2+2xz+z^2-2(xy-yz+xz)=2(x^2+y^2+z^2)=3^2+5^2+8^2-2*7=98-14=84x^2+y^2+z^2=42