Chapter 13 - Vector Geometry - 13.5 Planes in 3-Space - Exercises - Page 685: 42

$$(9,18,3).$$

The parametric equations of the line are given by $$ x=1+4t, \quad y=9t, \quad z=-1+2t.$$ Now, substituting in the equation of the plane $$ x-z=6\Longrightarrow 1+4t+1-2t=6\\ \Longrightarrow 2t=4\Longrightarrow t=2. $$ So the point of intersection is given by $$(9,18,3).$$