Equation of plane containing the intersection of two planes

morsel
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Homework Statement


Find an equation of the plane that passes through the point (-1,2,1) and contains the line of intersection of the planes x + y - z = 2 and 2x - y + 3z = 1




Homework Equations


Equation of a plane:
a(x-x0) + b(y-y0) + c(z-z0) = 0


The Attempt at a Solution


n1 = <1,1,-1>
n2 = <2,-1,3>
n1\timesn2 = <2,-5,-3> = <a,b,c>

Equation:
2(x+1) - 5(y-2) - 3(z-1) = 0
2x - 5y - 3z = -15

What am I doing wrong? Any help is appreciated.
 
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What makes you think you are doing anything wrong?
 
The solutions section in the back of my textbook says the answer is x - 2y + 4z = -1.
My textbook occasionally has wrong answers, though. So my answer is OK?
 
Looking at it quickly, it looks right to me. But if you want to be absolutely certain, you could determine a couple of points on the intersection of the planes by using their equations. Then if those two points are on your proposed plane the whole line of intersection must be and you are certain.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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