Substitute the values of x, y, and z ftom the parametric equations of the line into the equation of the plane and solve for t.
x = 4t
y = 4 + 5t
z = -4 -3t
Then 4t + 4 + 5t -4 -3t = - 10 ⇒ 6t = -10 ⇒ t = -5/3
By substituting the value of t = -5/3 back to the parametric equations of the line
we are getting the coordinates of the point of interest.
x = -20/3
y = 13/3
z = 1