(a) 证明方程组

没有唯一解，其中 $a$ 为常数。

(b) 已知 $a=-5$，证明 (a) 中的方程组是相容的，并从几何角度解释这一情形。

(c) 若改为 $a\ne-5$，证明 (a) 中的方程组是不相容的，并从几何角度解释这一情形。

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

系数矩阵的第二列是第一列的相反数，所以行列式必为零。

For the coefficient matrix,

because its second column is the negative of the first. The matrix is therefore singular. Hence the system does not have a unique solution for any $a$.

先用第一和第三式确定 $z$ 及 $x-y$，再检查这些值也满足第二式。令 $y$ 取任意值，可得到三平面的公共直线。

Subtracting the first equation from the third gives $5z=9$, so $z=9/5$. The first then gives $x-y=4-2(9/5)=2/5$. The second left-hand side becomes $2/5-3(9/5)=-5$, so for $a=-5$ all three equations hold whenever

Thus the system is consistent, and the three planes share a common line.

第一和第三式仍确定相同的 $z$ 及 $x-y$，因此第二式只能在 $a=-5$ 时成立。再检查各对平面的交线。

Any solution must satisfy $z=9/5$ and $x-y=2/5$, forcing $a=-5$. Thus there is no solution when $a\ne-5$.

The three normals are pairwise non-parallel and are all perpendicular to $(1,1,0)$. Each pair of planes therefore meets in a line parallel to $(1,1,0)$. The three lines are distinct because the planes have no common point, and form the edges of a triangular prism.