Chapter 12 - Vectors and the Geometry of Space - 12.5 Equations of Lines and Planes - 12.5 Exercises - Page 872: 47

$(\frac{2}{5},4,0)$

Substitute $x=\frac{t}{5}$, $y=2t$, $z=t-2$ in $10x-7y+3z+24=0$ $2t-14t+3t-6+24=0$ $18-9t=0$ $t=2$ Substitute $t=2$ in equation of line to get the point. $x=\frac{2}{5}$, $y=2(2)$, $z=2-2$ $x=\frac{2}{5}$, $y=4$, $z=0$ Hence, $(\frac{2}{5},4,0)$