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