First off we need to find the point where these 2 lines intersect. Since we are dealing with 3D instead of a regular 2D xy graph, the point of intersection for these 2 planes is actually a line!
So we just need to find the normal vectors of these 2 planes!
Put both of these planes in normal vector form:
1st plane : x+y+(k_)z=6 has normal vector= N_1= i+j+(k_)k (where k_ is the variable we have to find)
2nd plane: 2x-y-2z=-5 has normal vector= N_2= 2(i) - j - 2(z)
And we know the dot product between both normal vectors equals their magnitudes multiplied together also multiplied to cos(60 degrees) so it looks like:
(N_1)(N_2) = (Mag(N_1))(Mag(N_2))(cos(60 degrees)) (where Mag(N_1) is magnitude of N_1 vector and Mag(N_2) is magnitude of N_2 vector)
to solve for variable k_ set cos(60)= [(N_1)(N_2)]/[(Mag(N_1))(Mag(N_2))] so it is:
cos(60)= (1+2(k_))/(3(sqrt(2+(k_)^2))) and solve for k_ !!
so k_ = 0.675 using quadratic formula setting k_ as the x term you have to solve for!!