A circle S having centre $(\alpha ,\beta )$ intersects the parabola $y^2=4x$ at three points A, B and C such that normals at A, B and C are concurrent at (9, 6) and O is origin. Then which is/are CORRECT?

A) Slope of normals are 1, -3, 2

$\left|m_1\right|+\left|m_2\right|+\left|m_3\right|=1+3+2=6$

C) Negative slope of normal = -3

Magnitude = 3

$A(t_1^2,2t_1)=(1,-2), B(t_2^2,2t_2)=(9,6), C(t_3^2,2t_3)=(4,-4)$

equation of circle passing through $(1,-2)(9,6)$ and $(4,-4)$ is $S\equiv x^2+y^2-11x-3y=0$

$(\alpha ,\beta )=(\frac{11}{2},\frac{3}{2}), \alpha +\beta =\frac{14}{2}=7$

so passes through origin. circle is ${(x-\alpha )}^2+{(y-\beta )}^2=r^2$

equation of normal for parabola $tx+y=2t+t^3$
It passes through (9, 6) $9t+6=2t+t^3, t^3-7t-6=0, t=-1^{-1}|\frac{\begin{array}{cccc}1& 0& -7& -6\\ 0& -1& 1& 6\end{array}}{\begin{array}{cccc}1& -1& -6& \overline{)0}\end{array}}$

$t^2-t-6=0, t^2-3t+2t-6=0, t = 3, - 2$