Unique solution for parametric system of two equations The system is: 
$$x - 2y + z = 2\alpha \\
3(xy + xz + yz) = 3\alpha - 4$$
I have to find for which values of $\alpha$, the system has an unique solution.
I've tried to simplify both expressions, set $x,y,z$ to some special values, and even apply some well-known inequalities to check for validity.
Any hint will be greatly appreciated!
 A: Here is a "by hand" method to make some progress.
You have from the first equation $x+z=2(y+\alpha)$
The second equation can be rewritten as $(x+z)y+xz=\alpha-\frac 43=2y(y+\alpha)+xz$
Both equations are symmetric in $x,z$ so if there is a single solution we must have $x=z$
Hence $x=z=y+\alpha$ and $x^2=\alpha-\frac 43-2y(y+\alpha)=(y+\alpha)^2$
Now there is a quadratic in $y$ namely $$9y^2+12\alpha y+3\alpha^2-3\alpha+4=0$$ or $$(3y+2\alpha)^2=\alpha^2+3\alpha-4=(\alpha+4)(\alpha-1)$$This has a single root for $y$ only if $\alpha=-4$ or $\alpha =1$
This identifies possible values for $\alpha$ on the assumption that the solution is unique. However it remains to be checked whether the solutions obtained are unique for these values of $\alpha$. The method above can be used to obtain a quadratic for $x$ and $z$ with coefficients which are functions of $y$. The discriminant of this quadratic - a function of $y$ can be used to check for uniqueness.
A: I assume that the variables $x,y,z$ and the parameter $\alpha$ are supposed to be real. To begin with, one could paramtetrise the plane $x-2y+z=2\alpha$ e.g. as follows:
$$ \begin{align} x &= s + 2t, \\ y &= t - \alpha, \\ z &= -s. \end{align} $$
Substituting the above expressions for $x$, $y$ and $z$ in the equation of the quadric and simplifying, one obtains the equation
$$ -s^2 - 2st + 2t^2 - 2\alpha t = \alpha - \frac{4}{3}.$$
This can be rewritten into the equivalent form
$$ -\left(s^2 + 2st + t^2\right) + 3\left(t^2 - \frac{2\alpha}{3}t + \frac{\alpha^2}{9}\right) = \frac{\alpha^2}{3} + \alpha - \frac{4}{3},$$
or
$$ -\left(s+t\right)^2 + 3\left(t-\frac{\alpha}{3}\right)^2 = \frac{1}{3}(\alpha+4)(\alpha-1).$$
If $\alpha = -4$ or $\alpha = 1$, the above equation describes a pair of concurrent lines in the real plane with coordinates $(s,t)$. In all other cases, the equation represents a hyperbola in that plane. Therefore, the original system of equations has infinitely many solutions for all values of $\alpha$ (unless I made a mistake somewhere). 
A: Here is an alternative, shorter, but calculus based solution. The system has a unique solution only when the plane and the surface are tangent. When this occurs the two surfaces have the same normal direction. 
A normal to the plane is $(1,-2,1)$. The normal to $f=xy+xz+yz$ is given by the gradient,
$(\frac{f}{dx} ,\frac{f}{dy} ,\frac{f}{dz})=(y+z, x+z, x+y )$ so we have the system of equations
$$y+z=\beta$$
$$x+z=-2\beta$$ 
$$x+y=\beta.$$
Solving we have 
$$x=-\beta$$
$$y=2\beta$$
$$z=-\beta.$$
If we substitute these into the original equations we get,
$$-6\beta=2\alpha$$
$$-9\beta^2=3\alpha -4$$
which gives the equation for $\beta$,
$$9\beta^2-9\beta-4=0$$
which factors
$$(3\beta+1)(3\beta-4)=0$$
so $\beta=\frac{4}{3}, -\frac{1}{3}$ and 
$$\alpha=-3\beta=-4, 1$$
Which (thankfully) is the same obtained by the other methods.
We also have the unique solutions in these cases, 
$$(-\frac{4}{3}, \frac{8}{3}, -\frac{4}{3})$$ and 
$$(\frac{1}{3}, -\frac{2}{3}, \frac{1}{3})$$
