The circle $C_{1} : x^{2} + y^{2} = 3,$ with centre at O, intersects the parabola $x^{2} = 2y$ at the point P in the first quadrant. Let the tangent to the circle $C_{1}$ at P touch other two circles $C_{2}$ and $C_{3}$ at $R_{2}$ and $R_{3},$ respectively. Suppose $C_{2}$ and $C_{3}$ have equal radii $2\cdot \sqrt{3}$ and centres $Q_{2}$ and $Q_{3},$ respectively. If $Q_{2}$ and $Q_{3}$ lie on the y-axis, then $(1) Q_{2}Q_{3} = 12 (2) R_{2}R_{3} = 4\cdot \sqrt{6} (3)$ area of the triangle OR$_{2}R_{3}$ is $6\cdot \sqrt{2} (4)$ area of the triangle PQ$_{2}Q_{3}$ is $4\cdot \sqrt{2}$

The circle C₁ : x² + y² = 3, with centre at O, intersects the parabola x² = 2y at the point P in the first quadrant. Let the tangent to the circle C₁, at P touches other two circles C₂ and C₃ at R₂ and R₃, respectively. Suppose C₂ and C₃ have equal radii 2√{3} and centres Q₂ and Q₃, respectively. If Q₂ and Q₃ lie on the y-axis, then $C_1: x^2 + y^2 = 3 \ (\text{centre } O), \qquad \text{parabola } x^2 = 2y$ $P = C_1 \cap (x^2 = 2y), \text{ in the first quadrant}$ $C_2, C_3: \text{ radius } 2\sqrt{3}, \ \text{centres } Q_2, Q_3 \text{ on the } y\text{-axis}$ $\text{Which of the statements (1)-(4) are correct?}$ $1.\ P = C_1 \cap (x^2 = 2y); \text{ tangent at } P$ $2.\ Q_i = (0, q): \text{ distance from } Q_i \text{ to tangent} = 2\sqrt{3}$ $3.\ \text{find } R_2, R_3, \text{ then test } (1)\text{-}(4)$ $x^2 + \left(\dfrac{x^2}{2}\right)^2 = 3 \Rightarrow x^4 + 4x^2 - 12 = 0$ $(x^2 + 6)(x^2 - 2) = 0 \Rightarrow x^2 = 2 \Rightarrow P(\sqrt{2}, 1)$ $\sqrt{2}\,x + y = 3 \qquad (\text{tangent at } P)$ $Q_i = (0, q): \ \dfrac{|q - 3|}{\sqrt{3}} = 2\sqrt{3} \Rightarrow |q-3| = 6$ $q = 9 \text{ or } q = -3 \Rightarrow Q_2(0,9), \ Q_3(0,-3)$ $R_i = Q_i - \dfrac{(q_i - 3)}{3}(\sqrt{2}, 1)$ $R_2(-2\sqrt{2}, 7), \qquad R_3(2\sqrt{2}, -1)$ $Q_2Q_3 = |9 - (-3)| = 12 \ \text{✓ (1) is correct}$ $R_2R_3 = \sqrt{(4\sqrt{2})^2 + (-8)^2} = \sqrt{96} = 4\sqrt{6} \ \text{✓ (2) is correct}$ $[\triangle OR_2R_3] = \dfrac{1}{2}\left|(-2\sqrt{2})(-1) - (7)(2\sqrt{2})\right| = 6\sqrt{2} \ \text{✓ (3) correct}$ $[\triangle PQ_2Q_3] = \dfrac{1}{2}\cdot 12 \cdot \sqrt{2} = 6\sqrt{2} \neq 4\sqrt{2} \ \text{✗ (4) is false}$ $\boxed{\text{Correct statements: (1), (2), (3)}}$