Q:

Find 3D coordinate on same plane as given 3 coordinates

I have 4 coordinates on a plane and coordinates $A,B,C$ are known. Line $AB$ is 90 degrees to line $AD$. The distance between $A$ to $D$ is 3 meters. I am trying to find coordinate $D$.  
The coordinates are:
$A=(2,-4,7),\\  
B=(3,-5,4)  ,\\ 
C=(0,0,0)$
Plot of 4 coordinates

A:

You can find the normal to the plane by taking the cross product $\vec{CA}\times\vec{CB}=  \begin{align}
\begin{pmatrix}
           2 \\
           -4 \\
           7
         \end{pmatrix}
\times \begin{pmatrix}
           3 \\
           -5 \\
           4
         \end{pmatrix} = \begin{pmatrix}
           19 \\
           13 \\
           2
         \end{pmatrix}
  \end{align} = \colon \vec{n} 
$. 
Since we are in 3-dimensions, and we know $\vec{AB}$ and $\vec{AD}$ are both perpendicular to the normal and to each other, we can find the direction of 
$\vec{AD}$ by $ \vec{AB} \times \vec{n}= \begin{align}
\begin{pmatrix}
           1 \\
           -1 \\
           -3
         \end{pmatrix}
\times \begin{pmatrix}
           19 \\
           13 \\
           2
         \end{pmatrix} = \begin{pmatrix}
           37 \\
           -59 \\
           32
         \end{pmatrix}
  \end{align}
$. 
Hence, $\vec{OD}= \begin{pmatrix}
           2 \\
           -4 \\
           7
         \end{pmatrix}\pm\frac{3}{\sqrt{5874}}\begin{pmatrix}
           37 \\
           -59 \\
           32
         \end{pmatrix}$.

