The speed is the magnitude of the vector given by:
![v=\sqrt[]{v^2_x+v^2_y_{}}_{}](https://img.qammunity.org/2023/formulas/physics/college/djkmjjflnq8fak0ymnq0f3w96ngrkklbzw.png)
Plugging the components of the vector we have:
![\begin{gathered} v=\sqrt[]{(5)^2+(-12)^2} \\ v=\sqrt[]{25+144} \\ v=\sqrt[]{169} \\ v=13 \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/nle34do1slmy4aro2vdzggckb7qgcio3fy.png)
Now to find the direction we need to notice that the vector lies on the second quadrant of the coordinate plane, this means that the formula:

does not give the angle we are looking for but the supplementary angle.
Then the direction is given as:

Plugging the values given we have:

Therefore the spped of the plane is 13 and the direction is 113°