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