Three normals are drawn to the parabola  $y^2=4x$ from the point $(c, 0)$. These normals are real and distinct when

Any point on $y^2=4x\text{ is }\left(t^2,2t\right)\text{. Since }\frac{\mathrm{d}y}{ \mathrm{d}x}=\frac{2}{y}$ so ${\left.\frac{\mathrm{d}y}{ \mathrm{d}x}\right|}_{\left(t^2,2t\right)}=\frac{2}{2t}=\frac{1}{t}$

Hence equation of normal at $\left(t^2,2t\right)$ is

$Y-2t=-t\left(X-t^2\right)$

This passes through $(c,0)\text{ if }-2t=-t\left(c-t^2\right)$

$\Rightarrow t=0,t^2=c-2$

Thus the roots are real and distinct if $c > 2$ so $c = 3$ is the correct choice.