For what values of $\alpha$, $\beta$, $\gamma$ is there a unique plane passing through the points? Let $P=(1,0,2)$, $Q=(-2,1,1)$, $R=(\alpha,\, \beta,\, \gamma)$ be points in $\mathbb{R}^{3}$ where $\alpha,\,\beta,\,\gamma \in \mathbb{R}$. 
I need to determine for what values $\alpha$, $\beta$, $\gamma$ there exists a unique plane, call it $\Pi$ passing through the points $P$, $Q$, and $R$.
To that effect, I attempted to form $\overrightarrow{PQ} = \langle -3, 1, -1\rangle$ and $\overrightarrow{PR}=\langle \alpha - 1, \beta, \gamma-2 \rangle$, and then take the cross product to find my normal vector $\overrightarrow{n}=\langle a, b,c\rangle$.
I found that $\overrightarrow{n}=\langle \beta + \gamma -2, \, -\alpha + 3 \gamma - 5, \, -\alpha-3\beta + 1 \rangle$. 
So, the equation of a plane passing through the three points (using point $P$) is $$(\beta+\gamma-2)(x-1) + (-\alpha + 3\gamma - 5)(y-0) + (-\alpha - 3\beta + 1)(z-2) = 0. $$
However, from here, I am at a loss as to how to figure out what $\alpha,\,\beta,\,\text{and}\,\gamma$ are. Could somebody please tell me how to proceed?
Thank you in advance.
 A: HINT: the equation of the plane is given by
$$x=\alpha+s(1-\alpha)+t(-2-\alpha)$$
$$y=\beta-s\beta+t(1-\beta)$$
$$z=\gamma+s(2-\gamma)+t(1-\gamma)$$
and the two direction vectors don't have to be linear dependend
this means
$$[1-\alpha;-\beta,2-\gamma]\ne k[-2-\alpha;1-\beta;1-\gamma]$$
A: A general equation for a plane through three given points $(x_1, x_2, x_3), (y_1, y_2, y_3), (z_1, z_2, z_3)$ is
$$0 = \begin{vmatrix}x - x_1 & y - y_1 & z - z_1 \\ x_2 - x_1 & y_2 -
 y_1 & z_2 - z_1 \\ x_3 - x_1 & y_3 -
 y_1 & z_3 - z_1\end{vmatrix}$$
Plugging in your $P, Q, R$ gives
$$0 = \begin{vmatrix}x - 1 & y & z - 2\\ -3 & 1 & -1 \\ \alpha-1 & \beta & \gamma - 2\end{vmatrix} = (x-1)(\beta + \gamma - 2) + y(-\alpha + 3\gamma - 5) + (z-2)(-\alpha - 3\beta + 1)$$
The plane is not well defined if and only if
$$\beta + \gamma = 2$$
$$-\alpha + 3\gamma = 5$$
$$\alpha+ 3\beta = 1$$
This is a linear system with a solution
$$\begin{pmatrix}\alpha \\ \beta\\ \gamma \end{pmatrix} = \begin{pmatrix}-5\\ 2\\ 0\end{pmatrix} + t \begin{pmatrix}3\\ -1\\ 1\end{pmatrix}$$
where $t \in \mathbb{R}$.
A: Hint If one of the coefficients $(\beta+\gamma-2), (-\alpha + 3\gamma - 5)$ or $(-\alpha - 3\beta + 1)$ is non-zero, then you found the plane, and it is unique.
But what happens if $\overrightarrow{n}=\langle \beta + \gamma -2, \, -\alpha + 3 \gamma - 5, \, -\alpha-3\beta + 1 \rangle= \langle 0, 0, 0 \rangle$? 
Show that in this case there are more than one plane passing through the points.
A: Hint: Three points make a triangle, unless they are linear.
Also, there is a unique plane passing through a triangle.
A: Three points in space determine a unique plane unless they are colinear. Given three points $p_1,p_2,p_3 \in \mathbb R^3$, the determinant of their coordinates $$\begin{vmatrix}x_1&y_1&z_1\\x_2&y_2&z_2\\x_3&y_3&z_3\end{vmatrix}$$ is the volume of the parallepiped with edges defined by the origin and the three points. This volume collapses to $0$ iff the points are colinear, so the condition for there being a unique plane through your three points is $$\begin{vmatrix}\alpha&\beta&\gamma\\1&0&2\\-2&1&1\end{vmatrix}\ne0.$$
