(a) 求使方程组

有唯一解的 $k$ 的取值集合。

(b) 当该方程组没有唯一解时，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。

第二行减去第一行的两倍后，只剩第一列非零。沿这一行展开行列式。

Subtracting twice the first row from the second gives the coefficient determinant

Hence the system has a unique solution for $\boxed{k\in\mathbb R\setminus\{2\}}$.

也可直接消元。先从第二式与第一式的组合求 $x$，再依次求 $y,z$，同时检查分母为零的情形。

The second equation minus twice the first gives $(k-2)x=-2$. The third minus three times the first gives $4x+2y=0$. For $k\ne2$, these give $x=-2/(k-2)$ and $y=4/(k-2)$. The first equation then gives $3z=1-x-2y=1+3x$, hence $z=\frac13+x=(k-8)/[3(k-2)]$. This solution is unique. For $k=2$, the equation for $x$ is impossible.

在 $k=2$ 时比较第一式和第二式。再检查第三个平面的法向量是否与它们成比例。

When $k=2$, the first two planes have proportional normals. Twice the first equation would have right-hand side $2$, while the second has right-hand side $0$. Thus these planes are distinct and parallel. The third plane has a non-proportional normal and intersects each of them.