Find a vector parametric equation for the plane containing the points whose position vectors are
u = (1,−1, 0), v = (3, 0,−4) and w = (0, 2, 5).
You need the position vector of a point located in the plane and two independent direction vectors:
Let $\displaystyle \vec p = (x,y,z)$ denote the stationary vector of any point in the plane:
$\displaystyle \vec p = \vec u + s \cdot (\vec v - \vec u) + t \cdot (\vec w - \vec u)\ ,\ s,t\ \in \ \mathbb{R}$
Plug in the values you know and simplify:
$\displaystyle (x,y,z) = (1,-1,0) + s \cdot (2,1,-4) + t \cdot (-1,3,5)\ ,\ s,t\ \in \ \mathbb{R}$
This will give the equations:
$\displaystyle P:\left \lbrace \begin{array}{rcl}x&=&1+2s-t \\ y&=& -1+s+3t \\ z&=& -4s+5t \end{array} \right.$
A plane which contains three points will have the same coefficients as the normal vector the plane, which is also normal to any two vectors in the plane.
Two vectors in the plane can be found by
$\displaystyle \displaystyle \begin{align*} \overline{u\,v} &= \overline{u\,O} + \overline{O\,v} \\ &= -\overline{O\,u} + \overline{O\,v} \\ &= \overline{O\,v} - \overline{O\,u} \\ &= <3, 0, 4> - <1, -1, 0> \\ &= <2, 1, 4>\end{align*}$
and
$\displaystyle \displaystyle \begin{align*} \overline{u\,w} &= \overline{u\,O} + \overline{O\,w} \\ &= -\overline{O\,u} + \overline{O\,w} \\ &= \overline{O\,w} - \overline{O\,u} \\ &= <0, 2, 5> - <1, -1, 0> \\ &= <-1, 3, 5> \end{align*}$
So now to find the normal, you need to evaluate $\displaystyle \displaystyle \begin{align*} \overline{u\,v} \times \overline{u\,w} \end{align*}$.
$\displaystyle \displaystyle \begin{align*} \overline{u\,v} \times \overline{u\,w} &= \left|\begin{matrix}\phantom{-}i&j&k\\ \phantom{-}2&1&4 \\ -1&3&5\end{matrix}\right| \end{align*}$
Find our direction vectors
u,w = (0-1, 2 - -1, 5-0) = (-1, 3, 5)
u,v = (3-1, 0 - -1, -4 -0) = (2, 1, -4)
Now that we found the direction vectors, we need only apply one of the three points we already have to find our equation of the plane
(x, y, z) = (1, -1, 0) + r(2, 1, -4) + s(-1, 3, 5)