Final answer:
The parametric equations for the line of intersection of the planes are x = 3 - t, y = t, z = 0. The angle between the planes is approximately 72.5 degrees.
Step-by-step explanation:
To find the parametric equations for the line of intersection of the planes, we need to find the direction vector of the line. This can be done by taking the cross product of the normal vectors of the two planes. For the given planes (x + y + z = 3) and (x + 6y + 6z = 3), the normal vectors are (1, 1, 1) and (1, 6, 6) respectively. Taking the cross product of these vectors gives us (-5, 5, -5). The parametric equations for the line are: x = 3 - t, y = t, z = 0.
To find the angle between the planes, we can use the dot product of their normal vectors. The dot product of (1, 1, 1) and (1, 6, 6) is 1*1 + 1*6 + 1*6 = 13. The magnitudes of the vectors are sqrt(3) and sqrt(73). Using the formula cos(theta) = dot product / (magnitude1 * magnitude2), we can find the angle theta. Substituting the values, we get cos(theta) = 13 / (sqrt(3) * sqrt(73)). Taking the inverse cosine of this value gives us the angle between the planes as approximately 72.5 degrees.