We have to find the equation of a plane parallel to the line (3 + 5t, 2t, -2 + 3t) and passing through the point (-2, 3, 5)
First, we need to determine a line perpendicular to the given line. There are an infinite number of such lines.
Let the line be ax + by + cz = 0. As the dot product of perpendicular lines is 0.
5a + 2b + 3c = 0
Take any a, b, c that satisfy the relation: a = 1, b = -1 and c = -1 is an option. Therefore the line x - y - z = 0 is perpendicular to the given line. This is a normal to the vector.
The equation of the vector is x - y - z = d
As it passes through (-2, 3, 5)
-2 - 3 - 5 = d
=> d = -10
One plane that passes through (-2, 3, 5) and is parallel to (3+5t, 2t, -2+3t) is x - y - z + 10 = 0