Chapter 13 - Vector Geometry - 13.5 Planes in 3-Space - Exercises - Page 685: 41

$$(-2,3,12).$$

The parametric equations of the line are given by $$ x=-6t, \quad y=9t, \quad z=36t.$$ Now, substituting in the equation of the plane $$ z=12\Longrightarrow 36t=12\\ \Longrightarrow t=\frac{1}{3}. $$ So the point of intersection is given by $$(-2,3,12).$$