Solution:

(a) First a point on the line must be found. By setting z = 0, we will find the point where the line intersects the x-y plane.

x + y = 2   ->  4x + 4y = 8

3x - 4y = 6          3x - 4y = 6

7x = 14  -> x = 2  and y = 0
 

Thus the line intersects the x-y plane at the point ( 2, 0, 0).

The two normal vectors for the planes are < 1, 1, -1 > and < 3, -4, 5 >. The cross product of the two vectors will give a vector that is parallel to the line of intersection.

The symmetric equations for the line of intersection are:

(b) The angle of intersection can be found using the equation:

When the planes intersect they form two angles, one obtuse and one acute, whose sum is 180 degrees. The acute angle is 180 - 119 = 61.