## Thomas' Calculus 13th Edition

$x=1-5t; y=-2+3t; z=-1+4t$
Consider the standard equation of a plane passing through the point $(x_1,y_1,z_1)$ to be $a(x-x_1)+b(y-y_1)+c(z-z_1)=0$ The equation of a normal plane is: $x+3y-z=-4$ $x=3+2t; y=2t; z=t$ Then $(3+2t)+3(2t)-t=-4 \implies 7t=-7$ or, $t=-1$ and $x=3+2(-1); y=2(-1); z=-1$ Therefore, the P co-ordinates are: $(1,-2,-1)$ Thus, the parametric equations are: $x=1-5t; y=-2+3t; z=-1+4t$