Chapter 12 - Vectors and the Geometry of Space - 12.5 Equations of Lines and Planes - 12.5 Exercises - Page 872: 46

$(-4,-5,6)$

Substitute $x=t-1$, $y=1+2t$, $z=3-t$ in $3x-y+2z=5$ $3(t-1)-(1+2t)+2(3-t)=5$ $2-t=5$ $t=-3$ Substitute $t=-3$ in equation of line to get the point. $x=-3-1$, $y=1+2(-3)$, $z=3-(-3)$ $x=-4$, $y=-5$, $z=6$ Hence, $(-4,-5,6)$