1. Write equations of tangents to both conics

For the parabola $P: y^2=12x$, we have $4a=12 \Rightarrow a=3$.

A tangent to $y^2=4ax$ in slope form is $y=mx+\frac{a}{m}.$ Hence for $P$, any tangent is $y=mx+\frac{3}{m}, \qquad m\neq 0.$

For the ellipse $E:\frac{x^2}{6}+\frac{y^2}{3}=1,$ a line $y=mx+c$ is tangent iff $c^2=a^2m^2+b^2$ for $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. Here $a^2=6,\ b^2=3$, so tangent condition is $c^2=6m^2+3.$

Since the line is a common tangent, its intercept must satisfy both: $c=\frac{3}{m}, \qquad c^2=6m^2+3.$ Therefore, $\left(\frac{3}{m}\right)^2=6m^2+3.$ So $\frac{9}{m^2}=6m^2+3$ $9=6m^4+3m^2$ $2m^4+m^2-3=0.$ Let $u=m^2$. Then $2u^2+u-3=0=(2u+3)(u-1).$ Since $u=m^2\ge 0$, we get $m^2=1 \Rightarrow m=\pm 1.$

Thus the two common tangents are $T_1: y=x+3, \qquad T_2: y=-x-3.$

2. Check where these tangents meet the $x$-axis

For $y=x+3$, putting $y=0$ gives $x=-3.$ For $y=-x-3$, putting $y=0$ gives $x=-3.$ So both tangents meet the $x$-axis at $(-3,0).$ Hence Option C is true and Option D is false.

3. Find points of contact with the parabola

For parabola $y^2=4ax$, point of contact of tangent $y=mx+\frac{a}{m}$ is $\left(\frac{a}{m^2},\frac{2a}{m}\right).$ Here $a=3$.

• For $m=1$: $A_1=\left(\frac{3}{1^2},\frac{6}{1}\right)=(3,6).$
• For $m=-1$: $A_4=\left(\frac{3}{(-1)^2},\frac{6}{-1}\right)=(3,-6).$

4. Find points of contact with the ellipse

For ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, tangent at parameter $\theta$ is $\frac{xx_1}{a^2}+\frac{yy_1}{b^2}=1,$ and slope form can also be matched directly.

A convenient formula: for tangent $y=mx+c$ to the ellipse, point of contact is $\left(-\frac{a^2m}{c},\frac{b^2}{c}\right).$ Here $a^2=6,\ b^2=3$.

• For $T_1: y=x+3$, we have $m=1, c=3$: $A_2=\left(-\frac{6\cdot 1}{3},\frac{3}{3}\right)=(-2,1).$

• For $T_2: y=-x-3$, we have $m=-1, c=-3$: $A_3=\left(-\frac{6(-1)}{-3},\frac{3}{-3}\right)=(-2,-1).$

So the vertices are $A_1=(3,6),\quad A_2=(-2,1),\quad A_3=(-2,-1),\quad A_4=(3,-6).$

5. Compute area of quadrilateral $A_1A_2A_3A_4$

Using the shoelace formula in the order $A_1(3,6),\ A_2(-2,1),\ A_3(-2,-1),\ A_4(3,-6),$ we get $\text{Area}=\frac12\left|\sum x_i y_{i+1}-\sum y_i x_{i+1}\right|.$

First sum: $3\cdot 1+(-2)\cdot(-1)+(-2)\cdot(-6)+3\cdot 6=3+2+12+18=35.$

Second sum: $6\cdot(-2)+1\cdot(-2)+(-1)\cdot 3+(-6)\cdot 3=-12-2-3-18=-35.$

Therefore, $\text{Area}=\frac12|35-(-35)|=\frac12\cdot 70=35.$ So Option A is true and Option B is false.

6. Final conclusion

The correct options are: $\boxed{A,\ C}.$

This matches the stored correct answer.