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 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 $.