Here is a detailed, step-by-step solution to the problem, designed to be clear, educational, and comprehensive, as expected from an expert JEE Mathematics teacher.

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1. Understanding the Parabola and its Focus

The given equation of the parabola is $y^2 = 12x$. To understand its properties, we compare it with the standard form of a parabola which is $y^2 = 4ax$.

• Standard Parabola Equation: For a parabola of the form $y^2 = 4ax$, its vertex is at the origin $(0,0)$, its axis of symmetry is the x-axis, and its focus is located at $(a, 0)$.

By comparing $y^2 = 12x$ with $y^2 = 4ax$: $4a = 12$ $a = 3$ Therefore, the focus of this parabola is $S = (a, 0) = (3, 0)$.

A focal chord is defined as any chord that passes through the focus of the parabola. This means the line segment AB must pass through the point $(3, 0)$.

2. Equation of the Focal Chord AB

Let the slope of the focal chord AB be $m$. Since the chord AB passes through the focus $S(3, 0)$ and has a slope $m$, we can use the point-slope form of a linear equation: $y - y_1 = m(x - x_1)$ Substitute the coordinates of the focus $(x_1, y_1) = (3, 0)$: $y - 0 = m(x - 3)$ $y = mx - 3m$ To prepare for distance calculations, we rearrange this equation into the general form $Ax + By + C = 0$: $mx - y - 3m = 0$ This is the equation of the focal chord AB.

• Understanding the condition $m < \sqrt{3}$: This condition restricts the possible slopes of the chord. It ensures that the slope is finite and real. Importantly, for a focal chord to connect two distinct points A and B on the parabola, the slope $m$ cannot be $0$. If $m=0$, the equation becomes $y=0$ (the x-axis). The line $y=0$ intersects the parabola $y^2=12x$ only at $(0,0)$. For a focal chord to exist, it must connect two distinct points on the parabola, and it must pass through the focus $(3,0)$. The line $y=0$ does not pass through the focus $(3,0)$ unless we consider it as the $x$-axis itself, and in that case, $(3,0)$ is on the line, but it is not on the parabola. Thus, $m \neq 0$ is implicitly required for a true focal chord. Also, $m$ cannot be undefined (a vertical line), as that would not be included in $m < \sqrt{3}$.

3. Calculating the Length of the Focal Chord ($l$)

For a parabola $y^2 = 4ax$, the length $l$ of a focal chord with slope $m$ is given by a standard formula: $l = \frac{4a(1+m^2)}{m^2}$

• Tip for JEE: This formula is very useful and often appears in problems involving focal chords. It can be derived using parametric coordinates of the endpoints of the chord or using the polar equation of the parabola. Memorizing it saves valuable time.

We have already found $a=3$. Substituting this value into the formula for $l$: $l = \frac{4(3)(1+m^2)}{m^2}$ $l = \frac{12(1+m^2)}{m^2}$ This expression gives the length of the focal chord AB in terms of its slope $m$. Since $m \neq 0$, $l$ is a finite positive value.

4. Calculating the Distance of the Chord from the Origin ($d$)

We need to find the perpendicular distance $d$ from the origin $(0, 0)$ to the focal chord $mx - y - 3m = 0$.

• Distance of a Point from a Line Formula: The perpendicular distance $d$ from a point $(x_0, y_0)$ to a line $Ax + By + C = 0$ is given by: $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$

Here, the point is the origin $(x_0, y_0) = (0, 0)$. From the chord equation $mx - y - 3m = 0$, we identify $A=m$, $B=-1$, and $C=-3m$.

Substitute these values into the distance formula: $d = \frac{|m(0) - 1(0) - 3m|}{\sqrt{m^2 + (-1)^2}}$ $d = \frac{|-3m|}{\sqrt{m^2 + 1}}$ Since distance must be non-negative, we can write $|-3m| = 3|m|$: $d = \frac{3|m|}{\sqrt{m^2 + 1}}$ We need $d^2$ for the final expression. Squaring both sides: $d^2 = \left(\frac{3|m|}{\sqrt{m^2 + 1}}\right)^2$ $d^2 = \frac{(3m)^2}{(\sqrt{m^2 + 1})^2}$ $d^2 = \frac{9m^2}{m^2 + 1}$ Since $m \neq 0$, $d^2$ is a positive finite value.

5. Evaluating the Expression $l d^2$

Now we substitute the expressions we found for $l$ and $d^2$ into the target expression $l d^2$: $l d^2 = \left(\frac{12(1+m^2)}{m^2}\right) \times \left(\frac{9m^2}{m^2 + 1}\right)$ We can observe that several terms will cancel out: $l d^2 = \frac{12(m^2+1)}{m^2} \times \frac{9m^2}{(m^2+1)}$ Since $m \neq 0$, $m^2 \neq 0$. Also, $m^2+1$ is never zero for real $m$. Therefore, we can safely cancel these terms: $l d^2 = \frac{12 \times 9 \times (m^2+1) \times m^2}{m^2 \times (m^2+1)}$ $l d^2 = 12 \times 9$ $l d^2 = 108$

6. Critical Analysis and Conclusion

Based on the standard definitions and formulas for parabolas, focal chords, and distances, the value of $l d^2$ is consistently 108 for any valid slope $m$ (where $m \neq 0$ and $m < \sqrt{3}$). The condition $m < \sqrt