How to find the form of coordinates on coincident planes I have three simultaneous equations:


*$5x-2y+z=28$

*$3x-4y-5z=0$

*$-2x+3y+4z=2$
When I represent these equations as a 3x3 matrix and try to solve them, I find that $[x,y,z]=(1/0)*[0,0,0]$. I think this means the planes are coincident i.e. the same plane, therefore there are infinite solutions to the equations. My textbook gives the answer as $x=λ, y=2λ, z=8-λ$. I think this means that coordinates on the plane(s) will be of the form $(λ, 2λ, 8-λ)$, but I'm unsure how to reach this conclusion.
Could someone please explain how to find the form of the coordinates from the equations in the context of matrices?
Here is a picture of the textbook question:
Question
And here is a picture of my working:
Working
 A: No, when the determinant is zero, as is the case here, there need not be three coincident planes. Any two coincident planes, or one plane expressible as a linear combination of the other two (as we happen to have here),  will lead to a 1-parameter family of solutions.  
Your textbook's answer is wrong, or perhaps you looked up the wrong answer; the right answer starting from $x=\lambda$ is $y=-10-2\lambda, z = 8-\lambda$. 
Let's proceed as in Gauss-Jordan elimination via elementary operations, and see where we get in trying to solve these equations:
$$
\pmatrix{5&-2&1\\3&-4&-5\\-2&3&4} \pmatrix{x\\y\\z} = \pmatrix{28\\0\\2}
$$
$$
\pmatrix{1&-2/5&1/5\\3&-4&-5\\-2&3&4} \pmatrix{x\\y\\z} = \pmatrix{28/5\\0\\2}
$$
$$
\pmatrix{1&-2/5&1/5\\0&-14/5&-28/5\\0&11/5&22/5} \pmatrix{x\\y\\z} = \pmatrix{28/5\\-84/5\\66/5
}
$$
$$
\pmatrix{1&-2/5&1/5\\0&1&2\\0&11/5&22/5} \pmatrix{x\\y\\z}= \pmatrix{28/5\\6\\0
}
$$
$$
\pmatrix{1&0&1\\0&1&2\\0&0&0} \pmatrix{x\\y\\z}= \pmatrix{8\\6\\0
}
$$
and now we are "stuck" because there is no way to eliminate the $1$ and $2$ in the top two rows of the third column so there is no way to determine $z$.  So let's make $z$  some arbitrary number, say $\alpha$. Then that last equation says:
$$
x+\alpha = 8\\y +2\alpha = 6\\z=\alpha
$$
Or we could have insisted that $x=\lambda$ in which case $\alpha = 8-\lambda$ and 
$$
x = \lambda \\y =6-2\alpha = -10 - 2\lambda\\z=8-\lambda
$$
