Here is the rewritten solution, aiming for clarity, elaboration, and educational value for a JEE student.

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Detailed Solution for Triangle Area Calculation

The problem asks us to find the square of the area of a triangle $\mathrm{PQ}_1 \mathrm{Q}_2$. To do this, we need to:

1. Find the coordinates of point P.
2. Determine the equation of line L.
3. Find the coordinates of the centres $Q_1$ and $Q_2$.
4. Calculate the area of triangle $\mathrm{PQ}_1 \mathrm{Q}_2$.
5. Square the calculated area.

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1. Finding the Point of Intersection P

Key Concept: To find the points where two curves intersect, we must solve their equations simultaneously. The coordinates of any intersection point must satisfy both equations.

We are given the equations for the circle and the parabola:

1. Circle: $x^2 + y^2 = 3$
2. Parabola: $x^2 = 2y$

Step-by-step working: Since both equations contain $x^2$, the most efficient way to solve them simultaneously is by substitution. Substitute the expression for $x^2$ from the parabola equation ($x^2=2y$) into the circle equation: $(2y) + y^2 = 3$ Rearrange this equation to form a standard quadratic equation in $y$: $y^2 + 2y - 3 = 0$ Now, we solve this quadratic equation for $y$. We can factorize it: $(y+3)(y-1) = 0$ This yields two possible values for $y$: $y = -3 \quad \text{or} \quad y = 1$

Explanation and Selection: The problem statement specifies that point P lies in the first quadrant. In the first quadrant, both the $x$ and $y$ coordinates must be positive. Therefore, we must choose the positive value for $y$. So, the $y$-coordinate of P is $y_P = 1$.

Next, we substitute $y_P=1$ back into the parabola equation $x^2=2y$ to find the corresponding $x$-coordinate: $x^2 = 2(1)$ $x^2 = 2$ Taking the square root of both sides gives: $x = \pm \sqrt{2}$ Again, since P is in the first quadrant, its $x$-coordinate must be positive. So, the $x$-coordinate of P is $x_P = \sqrt{2}$.

Result: The point of intersection P in the first quadrant is $(\sqrt{2}, 1)$.

Tip for JEE: Always pay close attention to conditions like "first quadrant," "positive x-axis," etc. They are crucial for selecting the correct solution from multiple possibilities and avoiding incorrect subsequent calculations.

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2. Determining the Equation of Line L

Key Concept: If a line passes through a specific point, the coordinates of that point must satisfy the line's equation. This property allows us to determine unknown parameters in the line's equation.

We are given the general equation of line L: $\sqrt{2}x + y = \alpha$ We have just found that line L passes through point P$(\sqrt{2}, 1)$.

Step-by-step working: Substitute the coordinates of P$(\sqrt{2}, 1)$ into the equation of L to find the value of $\alpha$: $\sqrt{2}(\sqrt{2}) + 1 = \alpha$ $2 + 1 = \alpha$ $\alpha = 3$

Result: The complete equation of line L is $\sqrt{2}x + y = 3$. For distance calculations, it's often useful to write the line in the general form $Ax+By+C=0$: $\sqrt{2}x + y - 3 = 0$ Here, $A=\sqrt{2}$, $B=1$, and $C=-3$.

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3. Locating the Centres of Circles $C_1$ and $C_2$

Key Concepts:

1. Centre on y-axis: If a point lies on the y-axis, its x-coordinate is 0. Therefore, the centres $Q_1$ and $Q_2$ will have coordinates of the form $(0, k)$.
2. Condition for Tangency: A line is tangent to a circle if and only if the perpendicular distance from the centre of the circle to the line is exactly equal to the radius of the circle.

Given Information:

• The radius of circles $C_1$ and $C_2$ is $R = 2\sqrt{3}$.
• The centres $Q_1=(0, k_1)$ and $Q_2=(0, k_2)$ lie on the y-axis.
• Line L: $\sqrt{2}x + y - 3 = 0$ is tangent to both circles.

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

Step-by-step working: For line L, we have $A=\sqrt{2}$, $B=1$, $C=-3$. For a generic centre $Q=(0, k)$, we have $x_0=0$, $y_0=k$. The distance from $Q(0, k)$ to line L must be equal to the radius $R=2\sqrt{3}$.

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