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:
$A=(2,-4,7),\\
B=(3,-5,4) ,\\
C=(0,0,0)$
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, $\vec{OD}= \begin{pmatrix} 2 \\ -4 \\ 7 \end{pmatrix}\pm\frac{3}{\sqrt{5874}}\begin{pmatrix} 37 \\ -59 \\ 32 \end{pmatrix}$.