# Problems from Relative position of three planes

Study the relative position of the planes given by the following equations:$$\begin{array}{rrcl} \pi_1:&x+3y-5z-3 &=&0 \\ \pi_2:&2x-y+z&=&0 \\ \pi_3:&2x+6y-10z-7 &=&0\end{array}$$$See development and solution ### Development: Start by finding the status of the matrix $$M$$ and of the extended matrix $$M'$$: $$|M|=\left|\begin{matrix} 1 & 3 & -5 \\ 2 & -1 & 1 \\ 2 & 6 & -10 \end{matrix} \right|=0 \ \ \ \ \ \ \left|\begin{matrix} 1 & 3 \\ 2 & -1 \end{matrix} \right|=-7\neq0 \Rightarrow rank(M)=2$$$

$$|M'|=\left|\begin{matrix} 1 & 3 & 3 \\ 2 & -1 & 0 \\ 2 & 6 & 7 \end{matrix} \right|=-7 \Rightarrow rank(M')=3$$$Therefore there are some secant planes, so we need to determine if there also are parallels planes: $$\dfrac{1}{2}\neq\dfrac{3}{-1} \Rightarrow \pi_1 \mbox{ and } \pi_2 \mbox{ aren't parallel }$$$

$$\dfrac{1}{2}=\dfrac{3}{6}=\dfrac{-5}{-10}\neq\dfrac{3}{7} \Rightarrow \pi_1 \mbox{ and } \pi_3 \mbox{ are parallel }$$\$

Therefore the $$\pi_1$$ and $$\pi_3$$ are parallel and secant to the plane $$\pi_2$$

### Solution:

The planes $$\pi_1$$ and $$\pi_3$$ are parallel and secant to the plane $$\pi_2$$

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