解:依题意知:①曲线方程:z=x²+y²;
②未知线方程:x+y+z=1;
联立①②方程:x²+y²+x+y=1,化简(x+½)²+(y+½)²=3/2(附图);
由于两线方程均满足合并方程,那么所求点到原点距离即为⊙A:(x+½)²+(y+½)²=3/2上点到原地的距离,且圆心A(-½,-½),圆半径r=√6/2,OA=(0+½)²+(0+½)²=½;
则所求最长距离为:OB=OA+r=(1+√6)/2;
最短距离为:OC=OA-r=(1-√6)/2.
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