Step-by-step Solution:

1. Identify the properties of the parabola. The equation of the parabola is $y^2 = 4x$. Comparing this with the standard form $y^2 = 4ax$, we get $a=1$.

• The focus $S$ is at $(a, 0)$, which is $S = (1, 0)$.
• The vertex is at $(0, 0)$.
• The directrix is the line $x = -a$, which is $x = -1$.

2. Find the points of contact of the tangents. The equation of a tangent to the parabola $y^2=4x$ at a point $(t^2, 2t)$ is given by $ty = x + t^2$. The tangents are drawn from the point $P = (-2, 1)$. Since the tangent passes through $P$, we substitute its coordinates into the tangent equation: $t(1) = -2 + t^2$ $t^2 - t - 2 = 0$ Factoring the quadratic equation, we get: $(t-2)(t+1) = 0$ This gives two values for the parameter $t$: $t_1 = 2$ and $t_2 = -1$.

The points of contact, $P_1$ and $P_2$, are given by $(at^2, 2at)$. With $a=1$:

• For $t_1 = 2$: $P_1 = (1 \cdot 2^2, 2 \cdot 1 \cdot 2) = (4, 4)$.
• For $t_2 = -1$: $P_2 = (1 \cdot (-1)^2, 2 \cdot 1 \cdot (-1)) = (1, -2)$.

3. Analyze the geometric setup. We are given points $Q_1$ on line $SP_1$ and $Q_2$ on line $SP_2$ such that $PQ_1 \perp SP_1$ and $PQ_2 \perp SP_2$. This means that $Q_1$ is the foot of the perpendicular from point $P$ to the line containing segment $SP_1$, and $Q_2$ is the foot of the perpendicular from point $P$ to the line containing segment $SP_2$. This creates two right-angled triangles: $\triangle PQ_1S$ (right-angled at $Q_1$) and $\triangle PQ_2S$ (right-angled at $Q_2$). In both triangles, the segment $PS$ is the hypotenuse.

4. Calculate necessary lengths and vectors.

• Point $P = (-2, 1)$

• Focus $S = (1, 0)$

• Point $P_1 = (4, 4)$

• Point $P_2 = (1, -2)$

• The vector $\vec{SP} = P - S = (-2-1, 1-0) = (-3, 1)$. The length $PS = |\vec{SP}| = \sqrt{(-3)^2 + 1^2} = \sqrt{9+1} = \sqrt{10}$.

• The vector $\vec{SP_1} = P_1 - S = (4-1, 4-0) = (3, 4)$. The length $SP_1 = |\vec{SP_1}| = \sqrt{3^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = 5$.

• The vector $\vec{SP_2} = P_2 - S = (1-1, -2-0) = (0, -2)$. The length $SP_2 = |\vec{SP_2}| = \sqrt{0^2 + (-2)^2} = \sqrt{4} = 2$.

5. Determine the lengths $SQ_1$ and $PQ_1$. In the right-angled triangle $\triangle PQ_1S$, $SQ_1$ is the length of the projection of the segment $PS$ onto the line $SP_1$. $SQ_1 = \frac{|\vec{SP} \cdot \vec{SP_1}|}{|\vec{SP_1}|} = \frac{|(-3)(3) + (1)(4)|}{5} = \frac{|-9+4|}{5} = \frac{|-5|}{5} = 1$.

Using the Pythagorean theorem in $\triangle PQ_1S$: $PS^2 = PQ_1^2 + SQ_1^2$ $(\sqrt{10})^2 = PQ_1^2 + 1^2$ $10 = PQ_1^2 + 1$ $PQ_1^2 = 9 \implies PQ_1 = 3$.

6. Determine the lengths $SQ_2$ and $PQ_2$. Similarly, in the right-angled triangle $\triangle PQ_2S$, $SQ_2$ is the length of the projection of $PS$ onto the line $SP_2$. $SQ_2 = \frac{|\vec{SP} \cdot \vec{SP_2}|}{|\vec{SP_2}|} = \frac{|(-3)(0) + (1)(-2)|}{2} = \frac{|-2|}{2} = 1$.

Using the Pythagorean theorem in $\triangle PQ_2S$: $PS^2 = PQ_2^2 + SQ_2^2$ $(\sqrt{10})^2 = PQ_2^2 + 1^2$ $10 = PQ_2^2 + 1$ $PQ_2^2 = 9 \implies PQ_2 = 3$.

7. Evaluate options A, C, and D.

• A: $SQ_1 = 2$. This is FALSE. Our calculation shows $SQ_1=1$.
• C: $PQ_1 = 3$. This is TRUE.
• D: $SQ_2 = 1$. This is TRUE.

8. Calculate the distance $Q_1Q_2$. To find the length of $Q_1Q_2$, we can use the Law of Cosines in triangle $\triangle SQ_1Q_2$. We know the lengths of two sides, $SQ_1 = 1$ and $SQ_2 = 1$. We need the angle between them, $\angle Q_1SQ_2$, which is the same as the angle between the lines $SP_1$ and $SP_2$, i.e., $\angle P_1SP_2$.

We can find the cosine of this angle using the dot product of the vectors $\vec{SP_1}$ and $\vec{SP_2}$: $\cos(\angle P_1SP_2) = \frac{\vec{SP_1} \cdot \vec{SP_2}}{|\vec{SP_1}| |\vec{SP_2}|} = \frac{(3)(0) + (4)(-2)}{(5)(2)} = \frac{-8}{10} = -\frac{4}{5}$.

Now, apply the Law of Cosines to $\triangle SQ_1Q_2$: $Q_1Q_2^2 = SQ_1^2 + SQ_2^2 - 2(SQ_1)(SQ_2)\cos(\angle Q_1SQ_2)$ $Q_1Q_2^2 = 1^2 + 1^2 - 2(1)(1)(-\frac{4}{5})$ $Q_1Q_2^2 = 1 + 1 + \frac{8}{5} = 2 + \frac{8}{5} = \frac{10+8}{5} = \frac{18}{5}$.

$Q_1Q_2 = \sqrt{\frac{18}{5}} = \frac{\sqrt{18}}{\sqrt{5}} = \frac{3\sqrt{2}}{\sqrt{5}} = \frac{3\sqrt{10}}{5}$.

9. Evaluate option B.

• B: $Q_1 Q_2 = \frac{3\sqrt{10}}{5}$. This is TRUE.

Conclusion: Based on the calculations, options B, C, and D are TRUE.