Showing that a quadratic cannot have 3 roots using Determinants

Question

Consider the equation $px^2 + qx + r = 0$ and let us assume that a, b and c satisfies the above equation.

So, $$a^2p + aq + r = 0$$ $$b^2p + bq + r = 0 $$ $$c^2p + cq + r = 0$$ They can be represented using a matrix.

\begin{bmatrix}a^2&a&1\\b^2&b&1\\c^2&c&1\end{bmatrix} From the equation,the matrix sends a point $(p, q, r)$ to the origin $(0, 0, 0)$.

Now my book states the following:

There are two possibilities:

1. $(p,q,r)$ is at the origin
2. $(p,q,r)$ is some distinct point

$(1)$ is not possible as for that value of $(p,q,r)$, the equation is no more an equation. So the latter is the case. According to the book, for $(2)$ to be true, the matrix must be singular. I don't quite get this.

Answer 1

If the matrix equation

$$\begin{bmatrix}a^2&a&1\\b^2&b&1\\c^2&c&1\end{bmatrix}\begin{bmatrix}p\\q\\r\\\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$$

has a non-trivial solution $(p,q,r)\neq (0,0,0)$, then the matrix is singular by definition.

Answer 2

$$px^2+qx+r = 0$$ Imagine that $x$ has $3$ roots $a,b,c$ Representing in matrix $$\begin{bmatrix}a^2&a&1\\b^2&b&1\\c^2&c&1\end{bmatrix}\begin{bmatrix}p\\q\\r\\\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$$ Since the matrix product is $$\begin{bmatrix}0\\0\\0\end{bmatrix}$$ Then it's determinant must be zero $$\begin{bmatrix}a^2&a&1\\b^2&b&1\\c^2&c&1\end{bmatrix}$$ $$\Delta = (a-b)(a-c)(b-c)$$ $$ (a-b)(a-c)(b-c) = 0$$ This equation here mean the the quadratic polynomial is cursed to always have only $2$ roots $$ (a-b)(a-c)(b-c) = 0$$ If we image that $a,b$ are two of it's root and we solve the quadratic for $c$, the result of $c$ would be $c=a$ or $c=b$ $$ (a-b)(a-c)(b-c) = 0$$ Now if we take only $a$ as one of its root, related $b$ and $c$ linearly and solving for them, it would also turn out that the last root is either one of the first two