Answer
$x+y+z=4$
Work Step by Step
The parametric equations are:
$x=1+t; y=3-2t; z=0+t$
The normal vector is
$\lt 1,-2, 1 \gt \times \lt 1,1, -2 \gt=\lt -2(-2)-(1)(1), (1)(1)-(1) (-2), (-2)(1) \gt=\lt 3,3,3 \gt$
Now, the equation of a plane can be written as:
$3(x-0)+3(y-5) +3( z-(-1))=0$
or, $3x+3y-15+3z+3=0$
or, $x+y+z=4$
