Clearly, we can scale the coefficients of a given linear equation by any (non-zero) constant and the result is unchanged. Therefore, by dividing-through by $\sqrt{a_i^2+b_i^2}$, we may assume our equations are in "normal form":
$$\begin{align}
x \cos\theta + y \sin\theta - p &= 0 \\
x \cos\phi + y \sin\phi - q &= 0 \\
x \cos\psi + y \sin\psi - r &= 0
\end{align}$$
with $\theta$, $\phi$, $\psi$ and $p$, $q$, $r$ (and $A$, $B$, $C$ and $a$, $b$, $c$) as in the figure:
Then
$$C_1 = \left|\begin{array}{cc}
\cos\phi & \sin\phi \\
\cos\psi & \sin\psi
\end{array} \right| = \sin\psi\cos\phi - \cos\psi\sin\phi = \sin(\psi-\phi) = \sin \angle ROQ = \sin A$$
Likewise,
$$C_2 = \sin B \qquad C_3 = \sin C$$
Moreover,
$$D := \left|\begin{array}{ccc}
\cos\theta & \sin\theta & - p \\
\cos\phi & \sin\phi & - q \\
\cos\psi & \sin\psi & - r
\end{array}\right| = - \left( p C_1 + q C_2 + r C_3 \right) = - \left(\;p \sin A + q \sin B + r \sin C\;\right)$$
Writing $d$ for the circumdiameter of the triangle, the Law of Sines tells us that
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = d$$
Therefore,
$$\begin{align}
D &= - \left( \frac{ap}{d} + \frac{bq}{d} + \frac{cr}{d} \right) \\[4pt]
&= -\frac{1}{d}\left(\;ap + b q + c r\;\right) \\[4pt]
&= -\frac{1}{d}\left(\;2|\triangle COB| + 2|\triangle AOC| + 2|\triangle BOA| \;\right) \\[4pt]
&= -\frac{2\;|\triangle ABC|}{d}
\end{align}$$
Also,
$$C_1 C_2 C_3 = \sin A \sin B \sin C = \frac{a}{d}\frac{b}{d}\sin C= \frac{2\;|\triangle ABC|}{d^2}$$
Finally:
$$\frac{D^2}{2C_1C_2C_3} = \frac{4\;|\triangle ABC|^2/d^2}{4\;|\triangle ABC|/d^2} = |\triangle ABC|$$
No. If the matrix is singular, i.e. if $\det A=0$, the system has $0$ solution or an infinity of solutions.
The general criterion is based on the augmented matrix $[A|B]$:
Let $A$ be an $m\times n$ matrix, $r$ its rank, $B$ an $m\times 1$ matrix. The system of equations $AX=B$ has a solution if and only if $\operatorname{rank}A=\operatorname{rank}[A|B]$.
Furthermore the set of solutions is an affine subspace of $\mathbf R^n$ of dimension $n-r$.
Best Answer
You're asking for a visual proof of something that's not true.
There's a problem with talking about "the intersection point of $l_1$ and $l_2$" or "the positive side" of $l_3$. For example, if $a_1=a_2=b_1=b_2=c_1=c_2=0$, $l_1,l_2$ are just the whole of $\mathbb{R}^2$. If $a_3=b_3=0$ and $c_3>0$, everything is on the "positive side" of $l_3$.
However, let's ignore that. Consider $$y=0 \tag{$l_1$}$$ $$x=0 \tag{$l_2$} $$ and $$x+y+1=0 \tag{$l_3$}$$ Then we get $$M=\begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 0 & 1\end{pmatrix}.$$ Then $l_1$ is just the $x$-axis, $l_2$ is just the $y$-axis, and the intersection is the origin. This is on the positive side of $l_3$, since $1\cdot 0+1\cdot 0+1=1>0$. But $\det(M)=-1$.
However, if we just switch $l_1$ and $l_2$, we get the matrix $$\begin{pmatrix}1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix},$$ which has $\det(M)=1$. But we didn't change $l_1,l_2$, or $l_3$, and the origin (intersection of $l_1,l_2$) still lies on the positive side of $l_3$.