证明方程组

其中 $k$ 为常数，没有唯一解，并从几何上解释这一情况。

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。

先比较第一式和第三式的系数。左边成比例，但右边不成同样的比例，因此方程组无解。第二个平面的法向量与前两个不平行。

The left-hand side of the third equation is $-\frac32$ times that of the first, but $14\ne-\frac32\cdot5$. Adding $\frac32$ times the first equation to the third gives

which is impossible. Therefore the equations are inconsistent for every $k$, so they do not have a unique solution.

The first and third equations represent distinct parallel planes. The second has normal $(1,1,k)$, which is not proportional to $(14,-4,6)$ for any $k$. It intersects each of the parallel planes in a line, and there is no common point of all three.