Consider the system of linear equations in x,y,z: x+y+z=6, x+2y+3z=14, x+2y+az=b, where a,b are real constants.

The determinant of the coefficient matrix is $\begin{vmatrix}1&1&1\\1&2&3\\1&2&a\end{vmatrix}=a-3$, so a unique solution exists iff $a\ne3$ (A true). For $a=3$ the third equation minus the second gives $0=b-14$. So $b=15$ makes the system inconsistent (B false), while $b=14$ leaves the first two equations: subtracting, $y+2z=8$, so $y=8-2z$ and $x=z-2$; this is the line through $(-2,8,0)$ with direction $(1,-2,1)$ (C true). For $a=2,b=10$: equation 3 minus equation 2 gives $-z=-4$, so $z=4$, $y=0$, $x=2$; then $y+z=4\ne2=x$ (D false).