The given lines are
x = 1 + t
y = 2t
z = -3t
The plane is x + y - z = 3
If the lines intersect the plane at a point p, then the lines should satisfy the equation of the plane.
That is,
(1 + t) + (2t) - (-3t) = 3
1 + t + 2t + 3t = 3
1 + 6t = 3
6t = 2
t = 1/3
Therefore the coordinates of the point p are
x = 1 + 1/3 = 4/3
y = 2(1/3) = 2/3
z = -3(1/3) = -1
That is p (4/3, 2/3, -1)
Answer: The point p has coordinates (4/3, 2/3, -1)