Five points all lie on the conic

Let $F(x,y)=ax^2+bxy+cy^2+dx+ey+f$

we can solve the this system of linear equation:

$F(0,2)=F(-3,0)=F(1,5)=F(1,1)=F(-1,1)=0$ and get $$ \begin{bmatrix} a \\ b \\ c \\ d \\ e \\ f \\ \end{bmatrix} =\begin{bmatrix} (-7/6)f \\ (19/6)f \\ (-2/3)f \\ (-19/6)f \\ (5/6)f \\ f \\ \end{bmatrix}$$ where $f$ can not be $0$

Plugging each point into the given equation gives \begin{align*} 4 \, c + 2 \, e + f &= 0 \\ 9 \, a - 3 \, d + f &= 0 \\ a + 5 \, b + 25 \, c + d + 5 \, e + f &= 0 \\ a + b + c + d + e + f &= 0 \\ a - b + c - d + e + f &= 0 \end{align*} This system can be written in matrix form as $A\vec x=\vec 0$ where \begin{align*} A &= \left[\begin{array}{rrrrrr} 0 & 0 & 4 & 0 & 2 & 1 \\ 9 & 0 & 0 & -3 & 0 & 1 \\ 1 & 5 & 25 & 1 & 5 & 1 \\ 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & -1 & 1 & -1 & 1 & 1 \end{array}\right] & \vec x &= \left[\begin{array}{r} a \\ b \\ c \\ d \\ e \\ f \end{array}\right] \end{align*} Now, row reduce $A$ to obtain $$ \DeclareMathOperator{rref}{rref}\rref(A)= \left[\begin{array}{rrrrrr} 1 & 0 & 0 & 0 & 0 & \frac{7}{6} \\ 0 & 1 & 0 & 0 & 0 & -\frac{19}{6} \\ 0 & 0 & 1 & 0 & 0 & \frac{2}{3} \\ 0 & 0 & 0 & 1 & 0 & \frac{19}{6} \\ 0 & 0 & 0 & 0 & 1 & -\frac{5}{6} \end{array}\right] $$ This implies that our original system is equivalent to \begin{align*} a + \frac{7}{6} \, f &= 0 & b - \frac{19}{6} \, f&= 0 & c + \frac{2}{3} \, f&= 0 & d + \frac{19}{6} \, f&= 0 & e - \frac{5}{6} \, f&= 0 \end{align*}