Describe the region (point, line, or plane), if any, that is the intersection of these planes?

Question

The following equations represent planes in $R^3$:

$y-x-3z = 1$
$z+2y-3x=-4$
$3y-4x-2z=-3$

If put into augmented form, you get
-1 1 -3 | 1
-3 2 1 | -4
-4 3 -2 | -3

And the RREF of the system is
1 0 -7 | 6
0 1 -10 | 7
0 0 0 | 0

And finally, the result is $(6+7t, 7+10t, t)$ Now, my question is simple. How exactly would I know if the intersection is either a point, line, or plane?

Thanks in advance.

Answer 1

The solution space to the system of equations has as many dimensions as the number of free variables. Lines are one-dimensional and planes are two-dimensional, so...

Answer 2

Hint:

your solution is $$ \begin{bmatrix} x\\y\\z \end{bmatrix}= \begin{bmatrix} 6\\7\\0 \end{bmatrix}+t \begin{bmatrix} 7\\10\\1 \end{bmatrix} $$

and this the equation of a line.