We need the normals of the planes. Take the first plane and let's identify one point on the plane by plugging in values for x and y. Let x=y=0, then z=5, since (0,0,5) lies on the plane. Call this point P0 and call the general point on the plane P=(x,y,z). Now convert these points into position vectors P0 and P. The difference between these two vectors is a vector on the plane:
P-P0=<x,y,z-5>. Let the normal n be <a,b,c>, which must be perpendicular to all points on the plane, so the dot product of the normal and P-P0 must be zero:
<x,y,z-5>.<a,b,c>=0=ax+by+cz-5c, which is the equation of the plane.
Therefore, since we already know the equation of the plane is x-y+z-5=0, a=1, b=-1, c=1. The normal is therefore composed of the coefficients of the variables, hence n=<1,-1,1>.
Therefore:
the normals of these two planes are n1=<1,-1,1> and n2=<1,-3,0>.
The dot product of these normals contains the angle between them.
|n1|=√(12+(-1)2+12)=√3; |n2|=√(12+(-3)2+02)=√10.
n1.n2=√30cosθ, where θ is the angle between the planes.
<1,-1,1>.<1,-3,0>=1+3+0=4, so √30cosθ=4, cosθ=4/√30, θ=43° approx., so the planes are not parallel.
Since we've been asked to prove that the planes are parallel, clearly there must be an error in the question so let's assume that the second plane should have been x-3y+kz+6=0, so n2 <1,-3,k>. If the planes are parallel then the dot product of the normals would contain the angle zero. cos0=1.
We need the magnitude of the second normal: √(1+9+k2) so the dot product of the normals is:
√(30+3k2)=4+k,
30+3k2=16+8k+k2,
2k2-8k+14=0,
k2-4k+7=0, which has no real solution, so the planes can never by parallel, no matter what the coefficient of z is.
[If the planes intersect, they will do so along a line. If the planes were parallel it would not be possible for them to intersect.
x-y+z=5, x-3y=-6.
Let x=t, then t-3y=-6, 3y=t+6, y=2+⅓t; t=3y-6;
x-y+z=t-t/3-2+z=5, 2t/3+z=7, z=7-⅔t; t=3(7-z)/2.
The equation of the line of intersection can be written x=3y-6=3(7-z)/2 or (0,2,7)+λ(3,1,-2), where λ is a scalar quantity.]