Given Planes, $\pi_1:x+2y-z=0,\pi_2:-x+2y-3z-4=0$, and $\pi_3:x+6y-5z-4=0$. Determine nature and co-ordinates of intersection of planes.

Question

$\require{enclose}$ Consider the planes below

$\color{chocolate}{\pi_1:x+2y-z=0\\\pi_2:-x+2y-3z-4=0\\\pi_3:x+6y-5z-4=0}$

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a. Without solving the system, and by examining the normals, determine the nature of the intersection of the three planes. Include diagrams with your answers.

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b. Find the intersection of the planes.

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These were given as Assignment. I attached my solution. Can somebody check is if the solution is correct? And is it good?

NOTE: In the image, Coefficient of $x$ is $A$, $y$ is $B$ and $z$ is $C$. The equation of $\pi_1$ is numbered as ${\enclose{circle}{\kern .06em 1\kern .06em}}$, $\pi_2$ as ${\enclose{circle}{\kern .06em 2\kern .06em}}$ and $\pi_3$ as ${\enclose{circle}{\kern .06em 3\kern .06em}}.$

Answer 1

Your solutions are correct. A tip on how to check these answers for yourself: dot your solution line vector with the normals of the planes. If you get zero, it means they are perpendicular, so the line is in the plane (if at least one point on the line is in the plane).

Details: Your solution has vector $(-1,1,1)$, while your plane $\pi_1$ has normal vector $(1,2,-1)$. And note that $(-2,1,0)$ is on your line, but obviously that satisfies $x+2y-1z=0$.