Compute the equation of the plane through R with subspace generated by $\vec{u}$ and $\vec{v}$ . If $ P = (1, 2, 3), Q = (1,2,3),
R = (0, 1, -1),
\vec{u} = (0, 1, -1)
\text{ and } \vec{v} = (5, 1, 2)$, compute the equation of the plane through R with subspace generated by $\vec{u}$ and $\vec{v}$.

We have that $a(0,1,-1) + b(5,1,2) = (x,y,z)$. This must mean that
$$ 5b = x, a + b = y, -a + 2b = z $$
Adding the second and third expression, we get that
$$ 3b = y + z $$
Then minusing the $5*$first expressin by the expression in the previous line we get
$$ -3x + 5y + 5z = 0 $$
So the subpace generated by these two vectors is $\{(\frac{5y+5z}{3}, y, z) | y,z \in  \mathbb{R}\}$
I am not sure how to get the equation of the plane through R after this.
 A: To write the equation of a plane you need a normal vector $\vec n=(a,b,c)$ and a point
The equation is $ax+by+cz+d=0$
A normal vector is surely the cross product of $\vec n=\vec u\times \vec v=(3, -5, -5)$
The plane has equation $3x-5y-5z+d=0$ 
Knowing that it passes through $R(0,1,-1)$ we plug the coordinates and get
$3\cdot 0 - 5\cdot 1 -5\cdot (-1)1+d=0\to d=0$
So the requested plane has equation 
$3x-5y-5z=0$
Hope this helps
A: The problem is worded somewhat oddly, but I interpret it as asking for the plane through $R$ parallel to $\vec u$ and $\vec v$. 
An approach to these sorts of problems that’s somewhat similar to the way you’ve started is to work in homogeneous coordinates (i.e., use projective geometry). The homogeneous equation of a plane is $ax+by+cz+dw=0$. Just as in Euclidean geometry, three noncolinear points determine a plane, and we have three such points: $R$ and the points at infinity that correspond to the vectors $\vec u$ and $\vec v$. This gives us a system of three linear equations for the coefficients: $$\begin{align} b-c+d&=0 \\ b-c&=0 \\ 5a+b+2c&=0 \end{align}$$ or in matrix form, $$\begin{bmatrix}0&1&-1&1\\0&1&-1&0\\5&1&2&0\end{bmatrix}\begin{bmatrix}a\\b\\c\\d\end{bmatrix}=\begin{bmatrix}0\\0\\0\\0\end{bmatrix}.$$ The solution to this equation is the null vector of the matrix on the left, namely, $[-3:5:5:0]$. The Cartesian equation of this plane is thus $$3x-5y-5z=0.$$
A: The equation you got is a plane passing through the origin (a.k.a. a subspace), so if $R$ doesn't belong to this subspace, there is a translation of it that passes through $R$, in other words, you'll just need to sum something conveniently to the equation. 
Plug $R$ in your equation and you get $$-3\cdot 0+5\cdot 1-5\cdot 1=0,$$ so $R$ already is in the plane you found and there's nothing left to do.
