Answer:
The nonzero vector orthogonal to the plane is <-9,-8,2>.
Step-by-step explanation:
Consider the given points are P=(0,0,1), Q=(−2,3,4), R=(−2,2,0).


The nonzero vector orthogonal to the plane through the points P,Q, and R is


Expand along row 1.




Therefore, the nonzero vector orthogonal to the plane is <-9,-8,2>.