How to show whether 3 planes have a common line of intersection To show whether or not the 3 planes
$$x+y-2z=5\tag 1$$ 
$$x-y+3z=6 \tag2$$ 
$$x+5y-12z=12 \tag 3$$ all have a common line of intersection.
Can I do $(3)-(2)$ to get the line $6y-15z=6$ and $(1)-(2)$ to get the line $2y-5z=-1$ which is $6y-15z=-3$ , and say that as these aren't the same line, they don't have a common line of intersection?
Another thing that is confusing me is that if instead of eliminating $x$, I chose to eliminate $z$, I would get different lines in terms of $x$ and $y$. But how can I get the equations of two different lines by eliminating from the same pair of plane equations? There's only one line of intersection between any pair of planes, so surely I should only be able to get one unique line if I eliminate a variable from a pair of planes? 
Any help would be appreciated
 A: An elementary solution and notes to the OP
If $3$ planes have a unique common point then they don't have a common straight line. In order to see if there is a common line we have to see if we can solve the following system of equations:
$$
\begin{matrix}
x+y&-2z&=&5\\
x-y&+3z&=&6\\
\ x+5y&-12z&=&12.
\end{matrix}
$$
Adding the first equation to the second one we get $$2x+z=11.$$
Multiplying the second equation by $5$ and then adding it to the third equation we get $$3x+z=21.$$
These two equations have a unique solution:
$$x=10\text { and } z=-9.$$
Substituting these numbers back to any of the original equations we get $y=-23$.
So, the three planes have a unique common point; no common line exists.
Notes to the OP
If you take, say, $(1)$ and $(2)$ and eliminate one of the variables, say $x$ then you get an equation of a straight line in the plane $zy$.
This line is a perpendicular projection of the common line of  $(1)$ and $(2)$ to $yz$. If the planes $(1)$, $(2)$, and $(3)$ have a unique point then all of the possible eliminations will result in a triplet of straight lines in the different coordinate planes. 
By erecting a perpendiculars from the common points of the said line triplets you will get back to the common point of the three planes. 
I hope that this brief explanation helped you to understand better your own efforts.
A: HINT: Find normal vectors of the planes and check if three of them are linearly independent. If there is a common line for all the planes, then their normal vectors will lie within the same plane, therefore three of them will not be linearly independent.

In your specific case, 
$$
x+y-2z=5\implies \vec{n}_1 = \begin{bmatrix} 1\\ 1\\ -2\end{bmatrix}\\
x-y+3z=6\implies \vec{n}_2 = \begin{bmatrix} 1\\ -1\\ 3\end{bmatrix}\\
x+5y-12z=5\implies \vec{n}_3 = \begin{bmatrix} 1\\ 5\\ -12\end{bmatrix} 
$$
First, we need to check if the system of vector $\left\{\vec{n}_1, \vec{n}_2, \vec{n}_3 \right\}$ is clearly independent or not. The simplest way to do that is to compute rank of the matrix $\left[\vec{n}_1 \ \vec{n}_2\ \vec{n}_3 \right]$:
$$
\operatorname{rank}\Big(\left[\vec{n}_1 \ \vec{n}_2\ \vec{n}_3 \right]\Big) = \operatorname{rank}
\begin{pmatrix}
1 & 1& 1 \\
1 & -1 & 5 \\
-2& 3 & -12 
\end{pmatrix}
$$
If $\ \operatorname{rank}\!\left(\vec{n}_1 \ \vec{n}_2\ \vec{n}_3 \right)=2$, then the normal vectors are linearly dependent, yet still span a plane. 
Second, we need to find out if there is a point common for all three planes.
If you can find a common point and the rank of system of normal vectors is 3, then there is a line shared by all three planes.
A: first by solving 2 planes find y and z, where u have to consider z as t, hence u'l get parametric equation of y and z w.r.t t. now put this value in any plane which will give u parametric equation of x in terms of t only. Now substitute this values in any sphere, than u'll get quadratic equation in terms of t, if it is line than it will have two values, by which using which u can find the exact two points of intersection. Try it, it works.
tutorial is here and here.
