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 $$\pi: 3x-y-z=5$$.
See development and solution
Development:
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 $$rank (M) =3$$.
Solution:
The straight line $$r$$ and the plane $$\pi$$ are secant.