System of equations can be interpreted as intersection of $3$ planes in $3$-dimensional space

Question

I'm struggling with this question.

I have worked out that attempting to solve for the $x_i$ leads to a contradiction, and that $$\begin{vmatrix}1&4&6\\1&-2&1\\2&14&17\end{vmatrix}=0$$ So there is no solution for $x$. But what are the planes?

1. Three parallel planes
2. two parallel planes and one intersecting plane
3. three planes that intersect the other two but not at the same location Link

So I have narrowed down the answer to 3, 4, 5. Which one is it and how do we know?

Answer 1

Labelling the coefficient matrix's rows as $R_1,R_2,R_3$, we have $R_3=3R_1-R_2$ but $R_1$ and $R_2$ independent, but $3\cdot18-(-6)\ne-6$ so the third equation is not a linear combination of the first two. This means that

• the planes corresponding to $R_1$ and $R_2$ intersect in a line
• $R_3$'s plane is parallel to that line of intersection, but not parallel to the first two planes

Therefore the fifth answer is correct: there is no solution for $x$ even though none of the planes are parallel.