Consider the parabola $$y^2 = 4x$$. Let $$S$$ be the focus of the parabola. A pair of tangents drawn to the parabola from the point $$P = (-2, 1)$$ meet the parabola at $$P_1$$ and $$P_2$$. Let $$Q_1$$ and $$Q_2$$ be points on the lines $$SP_1$$ and $$SP_2$$ respectively such that $$PQ_1$$ is perpendicular to $$SP_1$$ and $$PQ_2$$ is perpendicular to $$SP_2$$. Then, which of the following is/are TRUE?

The given parabola is $$y^{2}=4x$$, which is of the form $$y^{2}=4ax$$ with $$a=1$$.
Its focus is therefore $$S(1,0)$$.

1. Tangents from the external point $$P(-2,1)$$

For $$y^{2}=4ax$$, the tangent at the point $$(t^{2},2t)$$ on the parabola is
$$ty=x+t^{2} \; -(1)$$

Since this tangent passes through $$P(-2,1)$$, substitute $$x=-2,\;y=1$$ in $$(1)$$:
$$t(1)= -2+t^{2}\;\Longrightarrow\;t^{2}-t-2=0$$
$$\Rightarrow\; t=\dfrac{1\pm\sqrt{1+8}}{2}=2,\,-1$$

Hence the two points of contact are
$$P_{1}(4,4)\quad (t=2),\qquad P_{2}(1,-2)\quad (t=-1)$$

2. Lines $$SP_{1}$$ and $$SP_{2}$$

$$S(1,0),\;P_{1}(4,4) \;\Rightarrow\; SP_{1}: \dfrac{x-1}{3}=\dfrac{y}{4}$$
Direction vector $$\mathbf{d}_{1}=(3,4)$$ with $$|\mathbf{d}_{1}|^{2}=3^{2}+4^{2}=25$$.

$$S(1,0),\;P_{2}(1,-2) \;\Rightarrow\; SP_{2}: x=1$$ (a vertical line).

3. Foot of the perpendicular from $$P$$ to $$SP_{1}$$  ($$Q_{1}$$)

The vector $$\overrightarrow{SP}=(-3,\,1)$$.
Scalar projection of $$\overrightarrow{SP}$$ on $$\mathbf{d}_{1}$$ is
$$\dfrac{\overrightarrow{SP}\cdot\mathbf{d}_{1}}{|\mathbf{d}_{1}|^{2}} =\dfrac{(-3)(3)+1\cdot4}{25}=\dfrac{-5}{25}=-\dfrac15$$

Therefore
$$Q_{1}=S+\left(-\dfrac15\right)\mathbf{d}_{1} =(1,0)+\left(-\dfrac15\right)(3,4) =\left(1-\dfrac35,\;0-\dfrac45\right) =\left(\dfrac25,\;-\dfrac45\right)$$

Lengths involving $$Q_{1}$$

$$SQ_{1}= \left|\,-\dfrac15\right|\,|\mathbf{d}_{1}|=\dfrac15\cdot5=1$$

Vector $$\overrightarrow{PQ_{1}}=(-2-\tfrac25,\;1+\tfrac45)=(-\tfrac{12}{5},\,\tfrac{9}{5})$$
$$PQ_{1}= \sqrt{\left(\tfrac{12}{5}\right)^{2}+\left(\tfrac{9}{5}\right)^{2}} =\sqrt{\dfrac{144+81}{25}}=\sqrt{\dfrac{225}{25}}=3$$

4. Foot of the perpendicular from $$P$$ to $$SP_{2}$$  ($$Q_{2}$$)

The line $$SP_{2}$$ is $$x=1$$. The perpendicular to a vertical line is horizontal, so $$y$$ remains $$1$$.

Hence $$Q_{2}=(1,1)$$.

Length $$SQ_{2}$$

$$SQ_{2}=\sqrt{(1-1)^{2}+(1-0)^{2}}=1$$

5. Distance $$Q_{1}Q_{2}$$

$$Q_{1}\left(\dfrac25,-\dfrac45\right),\;Q_{2}(1,1)$$
$$\overrightarrow{Q_{1}Q_{2}}=\left(1-\dfrac25,\;1+\dfrac45\right) =\left(\dfrac35,\;\dfrac95\right)$$
$$Q_{1}Q_{2}= \sqrt{\left(\dfrac35\right)^{2}+\left(\dfrac95\right)^{2}} =\sqrt{\dfrac{9+81}{25}} =\dfrac{\sqrt{90}}{5} =\dfrac{3\sqrt{10}}{5}$$

6. Verification of the options

Option A: $$SQ_{1}=2$$  →  False  ($$SQ_{1}=1$$)
Option B: $$Q_{1}Q_{2}=\dfrac{3\sqrt{10}}5$$  →  True
Option C: $$PQ_{1}=3$$  →  True
Option D: $$SQ_{2}=1$$  →  True

Therefore, the correct statements are:
Option B, Option C, and Option D.