Step-by-step Solution:

1. Find the intersection point P

We need to find the intersection of the circle ${C_1}:{x^2} + {y^2} = 3$ and the parabola ${x^2} = 2y$ in the first quadrant.

Substitute ${x^2} = 2y$ into the equation of the circle: $(2y) + {y^2} = 3$ ${y^2} + 2y - 3 = 0$ Factoring the quadratic equation: $(y+3)(y-1) = 0$ This gives two possible values for y: $y = 1$ or $y = -3$. Since the point $P$ is in the first quadrant, we must have $y > 0$. So, $y = 1$.

Now, we find the corresponding x-coordinate: ${x^2} = 2y = 2(1) = 2$ Since $P$ is in the first quadrant, $x > 0$. So, $x = \sqrt{2}$.

The intersection point is $P(\sqrt{2}, 1)$.

2. Find the equation of the tangent to C1 at P

The equation of the tangent to the circle ${x^2} + {y^2} = {r^2}$ at a point $({x_1}, {y_1})$ is given by $x{x_1} + y{y_1} = {r^2}$. For the circle ${C_1}:{x^2} + {y^2} = 3$ and the point $P(\sqrt{2}, 1)$, the tangent equation is: $x(\sqrt{2}) + y(1) = 3$ $\sqrt{2}x + y - 3 = 0$ Let's call this tangent line $L$. This line $L$ is also tangent to circles $C_2$ and $C_3$.

3. Find the centers Q2 and Q3

The centers ${Q_2}$ and ${Q_3}$ lie on the y-axis. Let a general center be $Q(0, k)$. The radii of ${C_2}$ and ${C_3}$ are given as $r = 2\sqrt{3}$. The distance from the center $Q(0, k)$ to the tangent line $L: \sqrt{2}x + y - 3 = 0$ must be equal to the radius.

The distance formula is: $d = \frac{|A{x_0} + B{y_0} + C|}{\sqrt{{A^2} + {B^2}}}$ $2\sqrt{3} = \frac{|\sqrt{2}(0) + 1(k) - 3|}{\sqrt{(\sqrt{2})^2 + 1^2}}$ $2\sqrt{3} = \frac{|k - 3|}{\sqrt{2 + 1}} = \frac{|k - 3|}{\sqrt{3}}$ $|k - 3| = 2\sqrt{3} \times \sqrt{3} = 2 \times 3 = 6$ This gives two possible values for $k$:

1. $k - 3 = 6 \implies k = 9$
2. $k - 3 = -6 \implies k = -3$

So, the centers of the circles ${C_2}$ and ${C_3}$ are ${Q_2}(0, 9)$ and ${Q_3}(0, -3)$ (the order is not important).

4. Evaluate the given options

A: ${Q_2}{Q_3} = 12$ The distance between ${Q_2}(0, 9)$ and ${Q_3}(0, -3)$ is: ${Q_2}{Q_3} = \sqrt{(0-0)^2 + (9 - (-3))^2} = \sqrt{0 + (12)^2} = 12$ So, option A is correct.

B: ${R_2}{R_3} = 4\sqrt{6}$ Let $T$ be the point where the tangent line $L$ intersects the y-axis. To find $T$, set $x=0$ in the equation of $L$: $\sqrt{2}(0) + y - 3 = 0 \implies y = 3$ So, $T$ is the point $(0, 3)$. Notice that $T$ is the midpoint of the segment ${Q_2}{Q_3}$: $(\frac{0+0}{2}, \frac{9-3}{2}) = (0, 3)$. Consider the right-angled triangle $\triangle T{Q_2}{R_2}$ (right angle at ${R_2}$). The sides are:

• Hypotenuse $T{Q_2}$: distance from $(0,3)$ to $(0,9)$ is $9-3=6$.
• Side ${Q_2}{R_2}$: radius of ${C_2}$, which is $2\sqrt{3}$.
• Side $T{R_2}$: distance from $T$ to the point of tangency ${R_2}$. By the Pythagorean theorem: $(T{Q_2})^2 = ({Q_2}{R_2})^2 + (T{R_2})^2$ $6^2 = (2\sqrt{3})^2 + (T{R_2})^2$ $36 = 12 + (T{R_2})^2$ $(T{R_2})^2 = 24 \implies T{R_2} = \sqrt{24} = 2\sqrt{6}$ Since $T$ is the midpoint of ${Q_2}{Q_3}$, by symmetry, $T{R_3} = T{R_2} = 2\sqrt{6}$. The points ${R_2}$, $T$, ${R_3}$ are collinear on the tangent line. Thus, the distance ${R_2}{R_3}$ is: ${R_2}{R_3} = T{R_2} + T{R_3} = 2\sqrt{6} + 2\sqrt{6} = 4\sqrt{6}$ So, option B is correct.

C: area of the triangle $O{R_2}{R_3}$ is $6\sqrt{2}$ The area of a triangle is given by $\frac{1}{2} \times \text{base} \times \text{height}$. Let the base be the segment ${R_2}{R_3}$. From option B, the length of the base is $4\sqrt{6}$. The height is the perpendicular distance from the origin $O(0,0)$ to the line containing ${R_2}{R_3}$, which is the tangent line $L: \sqrt{2}x + y - 3 = 0$. $\text{Height} = \frac{|\sqrt{2}(0) + 1(0) - 3|}{\sqrt{(\sqrt{2})^2 + 1^2}} = \frac{|-3|}{\sqrt{3}} = \sqrt{3}$ Note that this distance is also the radius of circle $C_1$, as expected. $\text{Area of } \triangle O{R_2}{R_3} = \frac{1}{2} \times (4\sqrt{6}) \times (\sqrt{3}) = 2\sqrt{18} = 2\sqrt{9 \times 2} = 2 \times 3\sqrt{2} = 6\sqrt{2}$ So, option C is correct.

D: area of the triangle $P{Q_2}{Q_3}$ is $4\sqrt{2}$ The vertices of the triangle are $P(\sqrt{2}, 1)$, ${Q_2}(0, 9)$, and ${Q_3}(0, -3)$. Let's take the base of the triangle to be the segment ${Q_2}{Q_3}$, which lies on the y-axis. The length of the base is ${Q_2}{Q_3} = 12$ (from option A). The height of the triangle with respect to this base is the perpendicular distance from point $P$ to the y-axis, which is the absolute value of the x-coordinate of $P$. $\text{Height} = |x_P| = |\sqrt{2}| = \sqrt{2}$ $\text{Area of } \triangle P{Q_2}{Q_3} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 12 \times \sqrt{2} = 6\sqrt{2}$ The option states the area is $4\sqrt{2}$. So, option D is incorrect.

Conclusion

The correct options are A, B, and C.