Relative positions of a straight line and a plane
Considering the straight line $r:\left\{\begin{array}{rcl} 2x-y+z-2&=&0 \\ x+y+2z-7&=&0\end{array}\right.$ determine its relative position to the plane $\pi: 3x-y-z=5$.
To find the relative position of the straight line $r$ with the plane $\pi$, we study the compatibility of the system formed by three equations: $$|M|=\left|\begin{matrix} 2 & -1 & 1 \\ 1 & 1 & 2 \\ 3 & -1 & -1 \end{matrix} \right|=-2-6-1-3-1+4=-9\neq0$$ Therefore the straight line and the plane are secant since $rank (M) =3$.
The straight line $r$ and the plane $\pi$ are secant.