Problems from Relative position of three planes

Study the relative position of the planes given by the following equations: $\begin{array}{rrcl} π_{1} : & x + 3 y - 5 z - 3 & = & 0 \\ π_{2} : & 2 x - y + z & = & 0 \\ π_{3} : & 2 x + 6 y - 10 z - 7 & = & 0 \end{array}$

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Development:

Start by finding the status of the matrix $M$ and of the extended matrix $M^{'}$ :

$| M | = | \begin{matrix} 1 & 3 & - 5 \\ 2 & - 1 & 1 \\ 2 & 6 & - 10 \end{matrix} | = 0 | \begin{matrix} 1 & 3 \\ 2 & - 1 \end{matrix} | = - 7 \neq 0 \Rightarrow r a n k (M) = 2$

$| M^{'} | = | \begin{matrix} 1 & 3 & 3 \\ 2 & - 1 & 0 \\ 2 & 6 & 7 \end{matrix} | = - 7 \Rightarrow r a n k (M^{'}) = 3$

Therefore there are some secant planes, so we need to determine if there also are parallels planes:

$\frac{1}{2} \neq \frac{3}{- 1} \Rightarrow π_{1} and π_{2} aren't parallel$

$\frac{1}{2} = \frac{3}{6} = \frac{- 5}{- 10} \neq \frac{3}{7} \Rightarrow π_{1} and π_{3} are parallel$

Therefore the $π_{1}$ and $π_{3}$ are parallel and secant to the plane $π_{2}$

Solution:

The planes $π_{1}$ and $π_{3}$ are parallel and secant to the plane $π_{2}$

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