Analytic Geometry: line parallel to a plane I'm trying to figure out how to solve this problem of Analytic geometry of the space:

I have a line described by the system:
$$\begin{cases}
2x-y+z-1 &=0\\
5x+3y-8& =0
\end{cases}$$
and I have to verify wether this line is parallel or not to the following plane:
$x-y+z+10=0$

I tried to make the reduced equations of the line and then swap the $z$ and $y$ I obtained into the plane equation.
Solved the equation with variable $x$ but then I got 3 solutions which don't seem to be the real point of intersection between the line and the plane.

Any idea of how should I solve this problem?
 A: The answers given are shorter, but you could also proceed in the following way: you are given a line (two equations) and a plane (one equation) and in $\mathbb{R}^3$ there are only three possibilities: the line is parallel to the plane and not a part of it, the line intersects the plane or the line is parallel to the plane and a part of it. 
Looking at your equations, we can consider them as one system of equations. These systems have either no, one or infinitely many solutions! This corresponds to the previous situations I have described. Using this approach, we find the following system of equations:
$$\begin{cases}
2x - y + z &= 1\\
5x + 3y    &= 8\\
x - y +z &= -10
\end{cases}$$
which you can put in an augmented matrix and row reduce. This gives you the following solution: $(x;y;z) =  (11; -15,66666 ; -35,66666)$ (where I used $;$ to separate the coordinates and $,$ to denote the decimal parts). Hence we find one solution, and so the line must intersect the plane.
To show you that the line does intersect, I have plotted this in GeoGebra using the 3D option and found an intersection piont $A$ with the coordinates I mentioned:

A: Note that a direction vector of the line under consideration is easy to obtain as we observe that it is nothing more than a vector that is perpendicular to the normal vectors $(2,-1,1), (5,3,0)$ of the two planes (The textbook you are referencing, I trust, shows you how to construct a vector perpendicular to two given nonzero vectors logically earlier as well as the general form of such a vector.). Calculation gives $(-3, 5, 11)$ as a direction vector of the line. If it is parallel to the second plane then the inner product of $(-3,5,11)$ and the normal vector $(1,-1,1)$ of the plane is $= 0$; but the inner product is $3$, so the line is not parallel to the plane.
Appendix: A vector $(x,y,z)$ is perpendicular to nonzero vectors $(x_{1},y_{1},z_{1}), (x_{2},y_{2},z_{2})$ iff $x_{1}x + y_{1}y + z_{1}z = x_{2}x + y_{2}y + z_{2}z = 0$. This is what happens in the calculation process. 
A: Another approach (without the use of matrices): 
Let me first describe some more general things. Suppose you have a plane in $\mathbb{R}^3$, then it contains at least some point $P = (x_0, y_0, z_0)$. Consider an arbitrary other point $Q = (x,y,z)$, then $Q$ is a point of the plane if the vector $$\vec{PQ} = Q - P = (x-x_0, y-y_0, z-z_0)$$ is a vector which lies in the plane. Suppose we have a vector $$\vec{n} = (a,b,c)$$ which is normal to this plane. This means that for any vector in the plane, we must have that $\vec{n}$ is orthogonal to it (by definition of the normal vector). We described all possible vectors in the plane (see the expression of $\vec{PQ}$, so we find that the equation of the plane is given by 
$$a(x-x_0) + b(y-y_0) + c(z - z_0) = 0. \quad \quad\quad\quad\quad(1)$$
Where does this expression comes from? Well, two vectors $(x_1, x_2, x_3)$ and $(y_1, y_2, y_3)$ are orthogonal if and only if 
$$x_1y_1 + x_2y_2 + x_3y_3 = 0 $$
so because of our description of vectors $\vec{PQ}$ in the plane and the definition of normal vector, we find the equation in formula (1). 
Now for your line: let us find the direction of it by applying the same trick as for the plane: we look for a vector which lies on this line and do so by finding two points on the line and by computing the corresponding vector. You can check that the points $K= (1,1,0)$ and $L= (-2,6,11)$ are points on the line. I have found these by setting $z = 0$ and then by trying some $x$ values in order to find nice $y$-values. Hence the vector 
$$\vec{KL} = L - K = (-3, 5, 11)$$
is a vector on the line you have. 
Now we look if this vector is orthogonal to the normal of our plane. Because of formula (1), the normal is given by $(1,-1,1)$ (by comparing the equation of the plane you are given by the equation in formula (1)). If the normal is orthogonal to the vector on the line, then the line must be parallel to the plane. If not, the line will intersect the plane. We compute the inner product and find:
$$1 \cdot (-3) + (-1) \cdot 5 + 1 \cdot 11 = -3 -5 + 11 = 3 \neq 0$$
hence the vector on the line is not perpendicular to the normal of the plane. So the line will intersect the plane.
A: Hint The line of intersection is in both planes, therefore its direction is perpendicular to the normal directions of both planes. 
The cross product gives you the direction of this line. Check if this direction is orthogonal to the normal vector of the third plane.
A: we have from the plane the vector $$[1;-1;1]$$ and from the straight line $$[-3/11;8/11;1]$$ and the dot product gives zero, the straight line is parallel to the plane
