If $|z|^2+\bar{A}z^2+A(\bar{z})^2+B\bar{z}+\bar{B}z+c=0$ represents a pair of intersecting lines.... find the value of $|A|$.

If the given equation defines two intersecting lines then we can translate them such that they intersect at the origin. (Replace $z$ in the equation by $z+p+iq$, and determine $p$ and $q$ such that there are no linear terms in $z$, $\bar z$ appearing. The constant term will then automatically be zero, or we would not have two lines.)

The new equation is $$z\bar z+\bar A z^2+A\bar z^2=0,\qquad A=a\, e^{i\alpha}\ne0\ .\tag{1}$$ The map $$T:\quad{\mathbb C}\to{\mathbb C}, \qquad z\mapsto w=e^{-i\alpha/2}\, z$$ is a rotation. It moves the two lines into the $w$-plane without changing the angle of the intersection. The equation there is obtained from $(1)$ by letting $z:=e^{i\alpha/2}\,w$, hence is $$w\bar w+a(w^2+\bar w^2)=0\ .$$ With $w=u+iv$ this means $(u^2+v^2)+2a(u^2-v^2)=0$, or $$v=\pm\sqrt{{2a+1\over2a-1}}\>u\ .$$ These are two lines symmetric to the $u$-axis. It follows that $$\tau:=\tan{\theta\over2}=\sqrt{{2a+1\over2a-1}}\ ,$$ so that $$\cos\theta={1-\tau^2\over1+\tau^2}=-{1\over2a}\ .$$ This implies $$|A|=a=-{1\over2\cos\theta}={1\over2\cos\theta'}\ ,$$ where $\theta'$ is the angle $<{\pi\over2}$ between the two lines.

Let $L$ be a straight line on the Argand plane. Since we can rotate $L$ to obtain a vertical straight line and the equation of a vertical straight line on the closed right half plane is given by $\Re(z)=x$ for some constant $x\ge0$, the equation of a general straight line is given by $\Re(\omega z)=x$ for some complex number $\omega$ on the unit circle. It follows that if the angle between two lines $\Re(\omega_1 z)=x_1$ and $\Re(\omega_2 z)=x_2$ is $\theta'$, then $|\cos\theta'|=|\Re(\omega_1\overline{\omega_2})|$.

Now suppose $z$ lies on the union of the two lines $\Re(\omega_1 z)=x_1$ or $\Re(\omega_2 z)=x_2$. Then $$(\omega_1 z+\overline{\omega_1 z}-2x_1) (\omega_2 z+\overline{\omega_2 z}-2x_2)=0.$$ Expand the LHS, we obtain $$2\Re(\omega_1\overline{\omega_2})|z|^2 +\omega_1\omega_2 z^2 +\overline{\omega_1\omega_2}\,\overline{z}^2 +pz+\overline{pz} +r=0,$$ where $p$ and $r$ are some constants. If this equation is equivalent to $$|z|^2 +\overline{A}z^2 +A\overline{z}^2 +\overline{B}z +B\overline{z} +c=0,$$ by comparing the coefficients of $|z|^2$ and $\overline{z}^2$ in both equations, we have $2\Re(\omega_1\overline{\omega_2})A=\overline{\omega_1\omega_2}$. Hence $$2|\cos\theta'||A|=2|\Re(\omega_1\overline{\omega_2})||A|=1.$$