## Thomas' Calculus 13th Edition

$(2,\dfrac{-20}{7},\dfrac{27}{7})$
Since, we are given that the parametric equations $x=2; y=3+2t; z=-2-2t$ Equation of plane $6x+3y-4z=-12$ Plug all the parametric equations $x=2; y=3+2t; z=-2-2t$ in equation of plane, we have $t=\dfrac{-41}{14}$ Now, the Parametric equations are: $x=2; y=3+2(\dfrac{-41}{14})=\dfrac{-20}{7}; z=-2-2(\dfrac{-41}{14})=\dfrac{27}{7}$ Hence, the line will meet at the point: $(2,\dfrac{-20}{7},\dfrac{27}{7})$