None of the normals are scalar multiples of each other, so none of your planes are parallel
If we multiply the third equation by 3 we get 3z in all three equations.
#2 - #1 gave me x+11y = 4
the new #3 - #2 gave me 6x - 18y = 3
solving those two equations in x and y gave me x = 5/4 and y = 1/4
subbing that back into one of the originals I got z = -15/4
So your 3 planes intersect in a point, namely (5/4,1/4,-15/4)
Use normal vectors to determine the intersection, if any, for each of the following groups of three planes. Give a geometric interpretation in each case and the number of solutions for the corresponding linear system of equations. If the planes intersect in a line, determine a vector equation of the line. If the planes intersect in a point, determine the coordinates of the point.
2x − 5y + 3z = −10
3x + 6y + 3z = −6
3x − 4y + z = −1
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