Why can't I find any solution of this three equations? $$2x+6y-z=10$$
$$3x+2y+2z=1$$
$$5x+4y+3z=4$$
I do not find any solution of this three equations.I know that if two equations or planes are parallel then we do't find any solution.But I can't understand what is the problem here.Are they parallel.and if they are parallel How can I understand that.please  help me.
 A: Note that $$
\det\left(\begin{array}{ccc}
2&6&-1 \\ 3&2&2 \\ 5 & 4 & 3
\end{array}\right)=0$$
So there is either no so solution or an infinte amount of solutions.
Now substract the second line from the first line three times, and substract it from the third line twice, tends to
$$\begin{array}{ccc}
-7x-7z &=& 7\\
-x-z &=& 2
\end{array}$$
Can you continue from this point?
A: According to Wolfram Alpha,
$$z = 2x + 6y - 10, z = -\dfrac{3x}{2} - y + \dfrac{1}{2}, z = -\dfrac{5x}{3} - \dfrac{4y}{3} + \dfrac{4}{3}$$
So there seems to be no solution for this problem.
If we augment 3-by-3 matrix with 3-by-1 matrix and determine reduced row echelon form of 3-by-4 matrix, then we obtain
$$\left[\begin{array}{rrr|r}
1 & 0 & 1 & 0\\
0 & 1 & -\frac{1}{2} & 0\\
0 & 0 & 0 & 1
\end{array}\right]$$
(Unfortunately, MathJax won't let me set the augmented matrix like this.)
So indeed, there is no solution for this problem.
A: There is no solution. Maybe try your luck with 
$$\tag{(2)+2(1)}7x+14y=21$$
$$\tag{(3)+3(1)}11x+22y=34$$
first.
A: If your three equations have solutions, they have to satisfy the followings :
$(2)+2\times (1)$ will give you
$$x+2y=3.$$
$(3)+3\times (1)$ will give you
$$x+2y=34/11.$$
However, these does not have any solution (these represent two parallel lines). 
So, there is no solution for the given three equations.
A: Note that 
$$\ det\begin{pmatrix}
2 & 6 & -1 \\ 
3 & 2  &  2 \\ 
5 & 4  &  3
\end{pmatrix}= 0$$ while $$rank \begin{pmatrix}
2 & 6 & -1 & 10\\ 
3 & 2  &  2 &   1\\ 
5 & 4  &  3 &   4
\end{pmatrix}=3$$
because $$\begin{vmatrix}
6 & -1  & 10 \\ 
2 &  2 &     1\\ 
4 &  3 &    4
\end{vmatrix} \neq 0$$ so the system has no solutions.
On the other hand reducing the augmented matrix
$$
\begin{pmatrix}
2 & 6 & -1 & 10\\ 
3 & 2  &  2 &   1\\ 
5 & 4  &  3 &   4 
\end{pmatrix}
$$
$$
\longrightarrow 
\begin{pmatrix}
2 & 6 & -1 & 10\\ 
0 & -2  &  1 &   -7\\ 
0 & 0  &  0 &   35
\end{pmatrix}
$$. Seeing the last row you realize that can't be possible.
