x2+y2+z2+2yz+2zx+2xy=0x^2+y^2+z^2+2yz+2zx+2xy=0x2+y2+z2+2yz+2zx+2xy=0
Let x = y = z – 1 = t
t2+t2+(t+1)2+2t(t+1)+2(t+1)t+2t2=0t^2+t^2+(t+1)^2+2t(t+1)+2(t+1)t+2t^2=0t2+t2+(t+1)2+2t(t+1)+2(t+1)t+2t2=0
4t2+t2+2t+1+2t2+2t+2t2+2t=04t^2+t^2+2t+1+2t^2+2t+2t^2+2t=04t2+t2+2t+1+2t2+2t+2t2+2t=0
9t2+6t+1=09t^2+6t+1=09t2+6t+1=0
(3t+1)2=0(3t+1)^2=0(3t+1)2=0
3t+1=0 ⟹ t=−133t+1=0 \implies t=-\frac{1}{3}3t+1=0⟹t=−31
Hence x=−13,y=−13,z=23x=-\frac{1}{3}, y=-\frac{1}{3}, z=\frac{2}{3}x=−31,y=−31,z=32
Therefore (−13,−13,23)(-\frac{1}{3},-\frac{1}{3},\frac{2}{3})(−31,−31,32) is the only intersection point, q.e.d. Answer: True.
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