$$\begin{bmatrix} 1 & 1 & 1\\ 1 & 2 & 4\\ 1 & 3 & 9\\ 1 & 4 & 16 \end{bmatrix} . \begin{pmatrix} c\\ b\\ a \end{pmatrix} = \begin{bmatrix} 6\\ 5\\ 7\\ 10 \end{bmatrix} $$
We have $$A = \begin{bmatrix} 1 & 1 & 1\\ 1 & 2 & 4\\ 1 & 3 & 9\\ 1 & 4 & 16 \end{bmatrix} \quad \text{and} \quad x = [c,b,a]^T$$ We are going to find a parabola $y = ax^2 + bx + c$ that fits the four data points $(1,6)$, $(2,5)$, $(3,7)$, and $(4,10)$ as closely as possible.
We have $y = [6,5,7,10]^T$. The solution to this least-squares problem is given by $$x^* = (A^T A)^{-1}A^T y$$ which gives the coefficients of the parabola that minimizes the sum of squared distances to the data points. We obtain $$x^* = \begin{pmatrix} c\\ b\\ a \end{pmatrix} = \begin{pmatrix} 8.5\\ -3.6\\ 1 \end{pmatrix} $$ Therefore, the minimizing parabola is given by $$y = 1x^2 - 3.6x + 8.5$$
Could you check my calculation? Because I'm not sure...