## University Calculus: Early Transcendentals (3rd Edition)

$(2,\dfrac{-20}{7},\dfrac{27}{7})$
The parametric equations are: $x=2; y=3+2t; z=-2-2t$ We have the equation of a plane: $6x+3y-4z=-12$ Now, we plug in all of the parametric equations into the above equation of a plane: $t=\dfrac{-41}{14}$ Thus, our parametric equations become: $x=2; y=3+2(\dfrac{-41}{14})=\dfrac{-20}{7}; z=-2-2(\dfrac{-41}{14})=\dfrac{27}{7}$ Hence, the line will meet at the point: $(2,\dfrac{-20}{7},\dfrac{27}{7})$