The given plane, , has normal vector
. Any plane parallel to this one has the same normal vector.
Let be any point in the plane we want. The plane contains the point (1, 1, -1), so an arbitrary vector in this plane is
and this is perpendicular to .
So the equation of the plane is
or equivalently,
First, we need to make the equation given to be in the form;
y=mx + b
where m is the slope and b is the intercept
8y - 3x = -8
Add 2x to both-side of the equation
8y = 3x - 8
Divide through the equation by 8
y = 3/8 x -1
comparing the above with y =mx + b
m = 3/8
slope of parallel equations are equal
From the options given the only equation with slope 3/8 is;
y= 3/8 x + 8
The above equation of the line is parallel to 8y - 3x = -8