Chapter 12 - Vectors and the Geometry of Space - 12.5 Equations of Lines and Planes - 12.5 Exercises - Page 872: 45

$(-2,6,3)$

Substitute $x=2-2t$, $y=3t$, $z=1+t$ in $x+2y-z=7$ $(2-2t)+2(3t)-(1+t)=7$ $2-2t+6t-1-t=7$ $t=2$ Substitute $t=2$ in equation of line to get the point. $x=2-2.2$, $y=3.2$, $z=1+2$ $x=-2$, $y=6$, $z=3$ Hence, $(-2,6,3)$