Find the angle between 2 planes

Plane 1: x+y+2z=1
Plane 2: r=(1,-2,2) + lambda(2,1,0) + alpha(0,1,-3) with lambda and alpha both real numbers

Plane 2 is

x = 1 + 2L
y = -2 + L + A
z = 2 - 3A

where I use L and A for lambda and alpha

Solve the system y = 0, z = 0 to get A = 2/3, L = 4/3, plugging into plane 2 gives x = 11/3 as x-intercept of plane 2.

Solve the system x = 0, z = 0 to get A = 2/3, L = -1/2, plugging into plane 2 gives y = -11/6 as y-intercept of plane 2.

Solve the system x = 0, y = 0 to get A = 5/2, L = -1/2, plugging into plane 2 gives z = -11/2 as z-intercept of plane 2.

Hence equation of plane 2 is x/(11/3) + y/(-11/6) + z/(-11/2) = 1. Multiply by 11 to see the equation of plane 2 is 3x - 6y - 2z = 11.

Now it is clear that a normal vector to plane 1 is [1,1,2] and a normal vector to plane 2 is [3,-6,-2].
Take the dot product of unit vectors in the directions of these to get

cos(t) = (1*3 - 1*6 - 2*2) / (sqrt(1^2+1^2+2^2) * sqrt(3^2+6^2+2^2)) = -1/sqrt(6) = -0.4082483

So, t, the angle between the planes is arccos(-0.4082483) = 1.99133 radians, or about 114.0948 degrees, or if you prefer the acute angle 180 - 114.0948 = 65.9052 degrees.

Better check this, because I didn't.

[(2,1,0) x (0,1,-3)] a normal vector to your second plane, where x is the cross product.

To get the cosine of the angle between your planes, that is the cosine of

(1,1,2) and [(2,1,0) x (0,1,-3)]

you divide their scalar product by the product of their norms.

So [(2,1,0) x (0,1,-3)] = (-3,6,2) whose norm is sqrt(9+36+4) = 7.

Then (1,1,2)*(-3,6,2) = 7 divided by 7 *sqrt(1,1,2) = 1/sqrt(6)

Angle = arccos(1/sqrt(6)) = 65.905157°