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


*

*Three parallel planes

*two parallel planes and one intersecting plane

*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?

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.
