Why the system of equation has a nontrivial solution? I am studying barycentric coordinate, and I encountered the following from my book:
Consider two parallel lines $u_1x + v_1y + w_1z = 0$ and $u_2x + v_2y + w_2z = 0$. Because
they are parallel, we know that the system
$$u_1x + v_1y + w_1z = 0$$
$$u_2x + v_2y + w_2z = 0$$
$$1 = x + y + z$$
has no solutions $(x, y,z)$. This is only possible when
$$
\begin{vmatrix}
u_1 & v_1 & w_1 \\ 
u_2 & v_2 & w_2 \\  
1 & 1 & 1
\end{vmatrix}=0
$$
However, this implies that the system of equations
$$u_1x + v_1y + w_1z = 0$$
$$u_2x + v_2y + w_2z = 0$$
$$0 = x + y + z$$
has a nontrivial solution! (Conversely, if the lines are not parallel, the determinant is nonzero,
and hence there is exactly one solution, namely $(0, 0, 0)$.)

I don't understand where is the implication from.
 A: A line can be expressed as a set of points with barycentric coordinates $[x:y:z]$ such that
$$ux+vy+wz=0$$
for some nonzero vector $(u,v,w) \in \Bbb{R}^3$. We also require $x+y+z=1$ for normalization.
Two lines $u_1x+v_1y+w_1z=0$ and $u_2x+v_2y+w_2z=0$ being parallel precisely means that there is no point with barycentric coordinates $[x:y:z]$ such that
$$\begin{cases} 
u_1x+v_1y+w_1z=0,\\
u_2x+v_2y+w_2z=0
\end{cases}$$
i.e. there is no $(x,y,z) \in \Bbb{R}^3$ such that
$$\begin{cases} 
u_1x+v_1y+w_1z=0,\\
u_2x+v_2y+w_2z=0,\\
x+y+z=1
\end{cases}$$
Therefore this linear system has no solution so the matrix of the system
$$A = \begin{bmatrix} u_1 & v_1 & w_1 \\ u_2 & v_2 & w_2 \\1 & 1 & 1\end{bmatrix}$$
has zero determinant. Namely, if $\det A\ne 0$, the matrix $A$ would be invertible so you could find a (unique) solution as $(x,y,z) = A^{-1}(0,0,1)$, which is a contradiction.
This implies that the system $A(x,y,z) = 0$ has a nontrivial solution because we can pick a nonzero vector $(x,y,z)$ in the nullspace of $A$.
Namely, note that $\mathbf{a} = (u_1,v_1,w_1)$ and $\mathbf{b} = (u_2,v_2,w_2)$ are linearly independent (otherwise they would determine the same line) so $\mathbf{a} \times \mathbf{b} \ne 0$. Now you can pick $(x,y,z) = \mathbf{a} \times \mathbf{b}$ and check that $A(x,y,z) = 0$. In fact, it turns out that $A(x,y,z) = \det A$ which we said is equal to $0$.
This isn't a contradiction with the fact that the lines are parallel since $(x,y,z)$ cannot be normalized to barycentric coordinates $[x,y,z]$ since $x+y+z = 0 $.
