Chapter 12 - Section 12.5 - Equations of Lines and Planes - 12.5 Exercises - Page 831: 21

$(4,-1,-5)$

For $L_1$: $x_1=2+t$, $y_1=3-2t$, $z_1=1-3t$ For $L_2$: $x_2=3+s$, $y_2=-4+3s$, $z_2=2-7s$ Solve the system of equations and we get $s=1$ and $t=2$ Substitute these values into the equations for $z_1=z_2$ Thus, $1-3(2)=2-7(1)$ $-5=-5$ ; which is true. So, the lines intersect. Points of intersections are: $P=(2+2,3-2(2),1-3(2))=(4,-1,-5)$