What is the geometric interpretation of this system of planes? $x-2y+z=-1$, $2x+y-3z=3$, $x+8y-9z=9$

Question

What is the geometric interpretation (coincidence/parallel/intersection) of the plane equations \begin{alignat*}{5} x - 2y + \phantom{0} z\; & =-1\\ 2x + \phantom{0} y - 3z\; & =\phantom{-}3\\ x + 8y - 9z\; & =\phantom{-}9 \end{alignat*}

Answer 1

Three planes can either intersect in a single point, or have no point in common, or have infinite points in common. You planes do not have a common point, but none of two planes are parallel. So we have the situation of any two planes that have a common line as an intersection. There are three such lines and they are parallel.

Answer 2

Each equation is a plane in $\mathbb{R}^3$. A vector orthogonal to the plane is given by the equation coefficients, e.g. $(1, -2, 1)$ for the first equation.

Two planes are parallel iff their orthogonal vectors are colinear. This is not the case here.

Three planes are in general position (i.e. their intersection is a point) iff their orthogonal vectors form a base of $\mathbb{R}^3$. This can be checked by computing the determinant of the matrix of the 3 vectors. I.e. here $\begin{pmatrix} 1 & -2 & 1 \\ 2 & 1 & -3 \\ 1 & 8 & -9 \end{pmatrix}$

Determinant is 0 (and not 18 as I wrongly calcuted by head) so intersections of the planes two by two are 3 parallel lines. As there is a solution $(0, 0, -1)$, the 3 lines are the same.

When two planes are not parallel, their intersection is a line, and you can compute a director vector of this line by the product of two orthogonal vectors of the planes.

Another way of checking that the intersections of the planes are parallel lines, is to compute 2 of the 3 director vectors this way, and to check that they are colinear. For the 2 first planes we find $(5, 5, 5)$, for the 2nd and 3rd planes we find $(15, 15, 15)$. They are colinear.