Intersection of two planes (without a given point)

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The two planes defined by the equations 2x-7y+5z+1=0 and x+4y-3z=0 intersect along a line PQ. To find the general equation of a plane through this line, one must ensure that the normal vector of the new plane is perpendicular to the direction vector obtained from the cross product of the two normals. Additionally, the new plane must contain a point on the intersection line. The general form of the plane can be expressed as a*x + b*y + c*z + d = 0, leading to a system of linear equations for the coefficients a, b, c, and d. An alternative approach involves eliminating one variable and expressing the others in terms of a parameter t.
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The Two Planes 2x-7y+5z+1=0 and x+4y-3z=0 intersect in a line PQ. What is the general equation of a plane through this line?

Have done cross product of the two normals to get vector form of the line, and parametric x,y and z. Would like advice on what to do next, Thanks
 
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Any other plane through that line would also have a normal that's perpendicular to the direction vector you found through the cross-product. It will also have to contain a point on the intersection line. Write the general plane as a*x+b*y+c*z+d=0 and translate those conditions into linear equations in a, b, c and d.
 
Another method would to be to eliminate one variable. Then just that variable be t, then find the other two variables in terms of t.

then r=(x,y,z)
 

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