The focus is $S=(1,0)$. Take a parametric point $P_1=(t^2,2t)$ on $y^2=4x$; its tangent is $ty=x+t^2$.

This tangent passes through $P=(-2,1)$: $t=-2+t^2$, i.e., $t^2-t-2=0\Rightarrow t=2,-1$.

So $P_1=(4,4)$ and $P_2=(1,-2)$.

Line $S{P}_1$: passes through $(1,0)$ and $(4,4)$, with slope $\frac{4}{3}$; equation $4x-3y-4=0$.

Line $S{P}_2$: passes through $(1,0)$ and $(1,-2)$; equation $x-1=0$.

$Q_1$ is the foot of perpendicular from $P(-2,1)$ on $4x-3y-4=0$. Using the standard formula, $\frac{x_1+2}{4}=\frac{y_1-1}{-3}=-\frac{4(-2)-3(1)-4}{16+9}=\frac{15}{25}=\frac{3}{5}.$

So $x_1=-2+\frac{12}{5}=\frac{2}{5}$, $y_1=1-\frac{9}{5}=-\frac{4}{5}$, hence $Q_1=\left(\frac{2}{5},-\frac{4}{5}\right)$.

$Q_2$ is the foot of perpendicular from $P(-2,1)$ on $x=1$, so $Q_2=(1,1)$.

Compute:

$S{Q}_1=\sqrt{{\left(1-\frac{2}{5}\right)}^2+{\left(\frac{4}{5}\right)}^2}=\sqrt{\frac{9}{25}+\frac{16}{25}}=1.$ (A is FALSE).

$P{Q}_1=\sqrt{{\left(\frac{2}{5}+2\right)}^2+{\left(-\frac{4}{5}-1\right)}^2}=\sqrt{\frac{144}{25}+\frac{81}{25}}=\sqrt{9}=3.$ (C is TRUE).

$S{Q}_2=\sqrt{(1-1{)}^2+(1-0{)}^2}=1.$ (D is TRUE).

$Q_1{Q}_2=\sqrt{{\left(1-\frac{2}{5}\right)}^2+{\left(1+\frac{4}{5}\right)}^2}=\sqrt{\frac{9}{25}+\frac{81}{25}}=\sqrt{\frac{90}{25}}=\frac{3\sqrt{10}}{5}.$ (B is TRUE).

$\therefore$ The correct options are (B), (C) and (D).