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!!!