Let $z=x+iy$ with $x,y\in\mathbb R$. Then $\bar z=x-iy,\qquad z\bar z=x^2+y^2.$

We are given $w=z\bar z+k_1z+k_2iz+\lambda(1+i),\qquad k_1,k_2\in\mathbb R.$

We must use the condition that $\operatorname{Re}(w)=0$ is a circle of radius $1$ in the first quadrant touching the line $y=1$ and the $y$-axis.

1. Write $\operatorname{Re}(w)$ and $\operatorname{Im}(w)$

First compute: $k_1z=k_1x+ik_1y,$ $iz=i(x+iy)=ix-y,$ so $k_2iz=-k_2y+ik_2x.$ Also, $\lambda(1+i)=\lambda+i\lambda.$ Hence $w=(x^2+y^2)+(k_1x-k_2y+\lambda)+i(k_1y+k_2x+\lambda).$

Therefore, $\operatorname{Re}(w)=x^2+y^2+k_1x-k_2y+\lambda,$ $\operatorname{Im}(w)=k_1y+k_2x+\lambda.$

2. Use the circle condition

The curve $\operatorname{Re}(w)=0$ is $x^2+y^2+k_1x-k_2y+\lambda=0.$ This is a circle.

For a circle $x^2+y^2+2gx+2fy+c=0,$ center is $(-g,-f)$ and radius is $\sqrt{g^2+f^2-c}$.

Comparing, $2g=k_1\Rightarrow g=\frac{k_1}{2},$ $2f=-k_2\Rightarrow f=-\frac{k_2}{2}.$ So center is $\left(-\frac{k_1}{2},\frac{k_2}{2}\right).$

Now the circle has radius $1$, lies in the first quadrant, touches the line $y=1$ and the $y$-axis ($x=0$).

A circle of radius $1$ touching $x=0$ must have center at distance $1$ from the $y$-axis, so $x ext{-coordinate of center}=1.$ A circle of radius $1$ touching $y=1$ and lying below it in the first quadrant must have center at distance $1$ from $y=1$, hence $y\text{-coordinate of center}=0.$

Thus center is $(1,0)$ and radius is $1$. So the circle is $(x-1)^2+y^2=1,$ or $x^2+y^2-2x=0.$

Comparing with $x^2+y^2+k_1x-k_2y+\lambda=0,$ we get $k_1=-2,\qquad k_2=0,\qquad \lambda=0.$

3. Equation of $\operatorname{Im}(w)=0$

Now $\operatorname{Im}(w)=k_1y+k_2x+\lambda=0.$ Substitute the values: $-2y=0\Rightarrow y=0.$

So the curve $\operatorname{Im}(w)=0$ is the $x$-axis.

4. Intersection of this line with the circle

Circle: $(x-1)^2+y^2=1.$ With $y=0$, $(x-1)^2=1$ so $x-1=\pm1\Rightarrow x=0,2.$

Hence the intersection points are $A=(0,0),\qquad B=(2,0).$ So $AB=2.$

Therefore, $30(AB)^2=30\cdot 4=120.$

5. Compare with stored answer

My derived answer is $120.$ The stored correct answer is $24$.

These do not match. The geometry and algebra both consistently give $AB=2$, hence $30(AB)^2=120$.

So I disagree with the stored answer.