## University Calculus: Early Transcendentals (3rd Edition)

$x=1-5t; y=-2+3t; z=-1+4t$
We have the equation of a plane: $x+3y-z=-4$ The parametric equations are: $x=3+2t; y=2t; z=t$ Now, the equation of a plane becomes: $(3+2t)+3(2t)-t=-4$ or, $7t=-7 \implies t=-1$ and $x=3+2(-1); y=2(-1); z=-1$ Thus, the co-ordinates of P are: $(1,-2,-1)$ We know that the standard equation of a plane passing through the point $(x_0,y_0,z_0)$ is written as: $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$ Then, for the point $(1,-2,-1)$, we have the parametric equations: $x=1-5t; y=-2+3t; z=-1+4t$