Hello, skeske1234!

Find the intersection of $\displaystyle P_1, P_2, P_3$ and describe geometrically.

. . $\displaystyle \begin{array}{cccc}P_1 & x+y-z&=&3 \\

P_2 & 2x-y+z&=&5 \\ P_3 & x-2y+2z&=&6 \end{array}$

We have: .$\displaystyle \left|\begin{array}{ccc|c}

1 & 1 & \text{-}1 & 3 \\ 2 & \text{-}1 & 1 & 5 \\ 1 & \text{-}2 & 2 & 6 \end{array}\right|$

$\displaystyle \begin{array}{c}\\ R_2 - R_1 \\ \\ \end{array} \left|\begin{array}{ccc|c} 1 & 1 & \text{-}1 & 3 \\ 1 & \text{-}2 & 2 & 2\\ 1 & \text{-}2 & 2 & 6 \end{array}\right| $

$\displaystyle \begin{array}{c} \\ \\ R_3-R_2 \end{array}

\left|\begin{array}{ccc|c}1 & 1 & \text{-}1 & 3\\ 1 & \text{-}2 & 2 & 2 \\ {\color{red}0} & {\color{red}0} & {\color{red}0} & {\color{red}4}\end{array} \right|$

The bottom row says: .$\displaystyle 0 \:=\:4$

There is no point of intersection.

Interpretation:

Taking the planes in pairs, the three lines of intersection are parallel.

The three planes enclose a triangular prism.

The end view looks like this:

Code:

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