For a plane in $\mathbb{R}^3$ with $\mathbf{r_0}$ a point that lies in the plane and $\mathbf{n}$ a vector normal to the plane, its equation can be given by:
$$
\mathbf{n}\cdot(\mathbf{r}-\mathbf{r_0})=0 \quad \mbox{where} \quad \mathbf{r}=(x,y,z)
$$
To solve your question, let us first rearrange the above equation:
$$
\mathbf{n}\cdot(\mathbf{r}-\mathbf{r_0})=0 \quad\Rightarrow\quad \mathbf{n}\cdot\mathbf{r}-\mathbf{n}\cdot\mathbf{r_0}=0 \quad\Rightarrow\quad \mathbf{n}\cdot\mathbf{r}=\mathbf{n}\cdot\mathbf{r_0}
$$
Substituting values, gives the following equation for the plane in Cartesian form:
$$
(1,7,-2)\cdot(x,y,z) = (1,7,-2)\cdot(3,1,6) \quad \Rightarrow \quad
x+7y-2z = -2
$$
If
$v = (4, 0 , -5) \tag{1}$
then clearly
$w_1 = (0, 1, 0) \tag{2}$
is orthogonal to $v$; so is
$w_2 = (5, 0, 4), \tag{3}$
so $w_1$ and $w_2$ must span the plane normal to $v$; I found $w_1$ and $w_2$ by simple inspection, that is, intelligent (I hope!) guesswork! The parametric equation of the plane is then
$(x, y, z) = sw_1 + tw_2 = s(0, 1, 0) + t(5, 0, 4) = (5t, s, 4t). \tag{4}$
We might also observe that $w_1$ is normal to $w_2$, so if we normalize $w_2$ to
$w_2' = (\dfrac{5}{\sqrt{41}}, 0, \dfrac{4}{\sqrt{41}}) \tag{5}$
we can express the plane in terms of the orthonormal pair $w_1$, $w_2'$:
$(x, y, z) = sw_1 + tw_2' = (\dfrac{5}{\sqrt{41}}t, s, \dfrac{4}{\sqrt{41}}t); \tag{6}$
and finally, we can always use the non-parametric vector form
$0 = v \cdot (x, y, z) = 4x - 5z. \tag{7}$
The vector $v$ given by (1) may be normalized to
$v' = (\dfrac{4}{\sqrt{41}}, 0, -\dfrac{5}{\sqrt{41}}) \tag{8}$
if so desired; then (7) becomes
$0 = v' \cdot (x, y, z) = \dfrac{4}{\sqrt{41}}x - \dfrac{5}{\sqrt{41}}z. \tag{9}$
Note that in the above I have been able to neglect inclusion of the point $P_0$ since here, as in the OP's $2$-dimensional example, $P_0 = 0$, the zero vector.
Hope this helps. Cheers,
and as always,
Fiat Lux!!!
Best Answer
In $\Re^3$, a plane can be characterized completely by a vector normal to the plane, and a point it goes through:
Imagine any plane. Note that every vector normal to it's surface is necessarily parallel to each other (even if you consider the normal pointing in the other direction, the lines all these normals live in are parallel). So, once you have a line you have your plane except for a translation in the direction of that normal, so you need to specify a point in the space that belongs to the plane to fully define it.
Now, in equations, a vector connecting to points in the plane must be perpendicular to $\vec{v}=(3,1,-6)$. Let's call this generic point in the plane $\vec{u}=(u_x,u_y,u_z)$ and for a second let $\vec{u}_0$ be the point in the plane that they are giving us (origin in this case), then $$0=\vec{v}\cdot(\vec{u}-\vec{u}_0)=3u_x+u_y-6u_z$$
Please, note that what needs to be perpendicular to $\vec{v}$ is not a point in the plane $\vec{u}$ but a vector contained (or parallel) to the plane itself and therefore we need to use $(\vec{u}-\vec{u}_0)$ in the dot product above. Since the origin is contained in the plane in your case, then the distinction is hard to grasp but it is very important to understand.