I unfortunately made an error in my placement of the largest circle, so I am going to work the problem again, from start to finish, using the correct location.

Let's begin by placing the two smaller circles in the plane as follows:

Now, the center of the largest circle must satisfy the nonlinear system:

\(\displaystyle (x+5)^2+(y+5)^2=14^2\)

\(\displaystyle (x+5)^2+(y-7)^2=16^2\)

Solving this system, and taking the solution which places the largest circle to the right of the others, we obtain:

\(\displaystyle (x,y)=\left(\frac{7}{2}\sqrt{15}-5,-\frac{3}{2}\right)\)

And so we now place the largest circle in the plane:

The vertical tangent line (tangent to the largest circle) will be:

\(\displaystyle x=\frac{7}{2}\sqrt{15}-14\)

The line tangent to the circle of radius 7 will have a slope of:

\(\displaystyle m=\frac{-\dfrac{3}{2}+5}{\dfrac{7}{2}\sqrt{15}-5-(-5)}=\frac{1}{\sqrt{15}}\)

The tangent point is obtained through:

\(\displaystyle \frac{7-y_1}{-5-x_1}=-\sqrt{15}\)

\(\displaystyle (x_1+5)^2+(y_1-7)^2=7^2\)

This gives us the tangent point:

\(\displaystyle \left(x_1,y_1\right)=\left(-\frac{13}{4},\frac{7}{4}(4-\sqrt{15})\right)\)

And so this tangent line has the equation:

\(\displaystyle y=\frac{1}{\sqrt{15}}x+7-\frac{23}{\sqrt{15}}\)

For the line tangent to the smallest circle, it's slope is:

\(\displaystyle m=\frac{-\dfrac{3}{2}-7}{\dfrac{7}{2}\sqrt{15}-5-(-5)}=-\frac{17}{7\sqrt{15}}\)

The tangent point is obtained through:

\(\displaystyle \frac{-5-y_1}{-5-x_1}=\frac{7\sqrt{15}}{17}\)

\(\displaystyle (x_1+5)^2+(y_1+5)^2=5^2\)

This gives us the tangent point:

\(\displaystyle \left(x_1,y_1\right)=\left(-\frac{75}{32},\frac{35}{32}\sqrt{15}-5\right)\)

And so this tangent line has the equation:

\(\displaystyle y=-\frac{17}{7\sqrt{15}}x+\frac{817}{224}\sqrt{\frac{3}{5}}-5\)

Here is a plot including the three tangent lines:

We then find that the length of the vertical side of the triangle whose area we are finding is:

\(\displaystyle \left(\frac{21}{2}-\frac{37}{\sqrt{15}}\right)-\left(\frac{10067}{224\sqrt{15}}-\frac{27}{2}\right)=\frac{1}{672}\left(16128-3671\sqrt{15}\right)\)

The area of the triangle whose vertices are at the centers of the circles is:

\(\displaystyle A_V=\frac{1}{2}(7-(-5))\left(\frac{7}{2}\sqrt{15}-5-(-5)\right)=21\sqrt{15}\)

Also:

\(\displaystyle A_V=k(12)^2\implies k=\frac{7\sqrt{15}}{48}\)

Hence, the area we want is:

\(\displaystyle A=\frac{7\sqrt{15}}{48}\left(\frac{1}{672}\left(16128-3671\sqrt{15}\right)\right)^2=\frac{154085333\sqrt{15}-592058880}{1032192}\approx4.56412\)

I sure hope I didn't make any other errors!!