Answer
$x=1-5t; y=-2+3t; z=-1+4t$
Work Step by Step
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:
$x=1-5t; y=-2+3t; z=-1+4t$
