Find parametric equations of the plane that is parallel to the plane 3x + 2y - z = 1 and passes through the point P(l, 1, 1).

Question

Find parametric equations of the plane that is parallel to the plane $$3x + 2y - z = 1$$ and passes through the point $$P(1, 1, 1)$$.

How would one approach this problem?

EDIT:

The normal vector is $$(3,2,-1)$$ So I was able to get two vectors $$(0,1,-1) $$ And $$(2,0,-2)$$

But what would be the translation of the two vectors?

Answer 1

For two parallel planes, the normal vector is th same, so the new plane has normal vector $\vec{n}=(3,2,-1)$, and plane have to passes through the point $$P(1, 1, 1)$$ then the equation of the plane is $$3(x-1)+2(y-1)-(z-1)=0$$ and

Answer 2

You can interpret $$ 1 = 3x + 2y-z = (3,2,-1)^T \cdot (x,y,z)^T \iff \\ d := \frac{1}{\lVert (3,2,-1)^T\rVert} = \frac{(3,2,-1)^T}{\lVert (3,2,-1)^T\rVert} \cdot (x,y,z)^T \iff \\ d = n \cdot v $$ as equation of a plane with a unit normal vector $n$ and signed distance $d$ to the origin.

A parallel plane differs by $d$.

Finally choose $d$ such that $v = (x,y,z)=(1,1,1)$ satisfies the above equation: \begin{align} d &= \frac{(3,2,-1)^T}{\lVert (3,2,-1)^T\rVert} \cdot (1,1,1)^T \\ &= \frac{4}{\sqrt{14}} \end{align} So $$ 3x + 2y - z = 4 $$ is the sought parallel plane.