Answer: The line passing through the point with vector 2a + 3b and parallel to c can be written as:
r = (2a + 3b) + t c
where t is a scalar parameter.
To find where this line intersects the plane, we can substitute the line equation into the plane equation and solve for t and s:
a - b + t(a + b - c) + s(a + c - b) = (2a + 3b) + t c
Expanding both sides and collecting terms, we get:
[(1 + t + s) a + (-1 - t + s) b - tc + ts c] = (2a + 3b)
Equating the coefficients of a, b, and c on both sides, we get the following system of equations:
1 + t + s = 2
-1 - t + s = 3
-t + ts = 0
Solving for t and s, we get:
t = -1
s = 1
Substituting t = -1 into the line equation, we get:
r = (2a + 3b) - c
Therefore, the position vector of the line where it intersects the plane is 2a + 3b - c.