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
The three lines are concurrent iff the system of equations is consistent. This will be the case when the augmented matrix $$A = \left[\begin{array}{cc|c}a_1&b_1&-c_1\\a_2&b_2&-c_2\\a_3&b_3&-c_3\end{array}\right]$$ has the same rank as the $3\times2$ coefficient matrix. The latter has rank at most 2, so a necessary condition is that $\det A=0$. Multiplying a column by $-1$ changes the sign of the determinant, but we only care about its vanishing, hence we can also use the matrix in your question.
In order to understand this in terms of normals, it might be helpful to make the equations homogeneous by moving to $\mathbb R^3$. The equation $a_ix+b_iy+c_iz=0$ describes a plane through the origin with normal $(a_i,b_i,c_i)^T$. For the three planes to intersect somewhere other than the origin, the homogeneous system’s coefficient matrix must be rank-deficient, which in turn is equivalent to its determinant vanishing. We connect these planes back to lines in $\mathbb R^2$ by intersecting them with the plane $z=1$. (This is one of the standard models of the real projective plane.)