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).
![\overrightarrow {PQ}==](https://tex.z-dn.net/?f=%5Coverrightarrow%20%7BPQ%7D%3D%3C-2-0%2C3-0%2C4-1%3E%3D%3C-2%2C3%2C3%3E)
![\overrightarrow {PR}==](https://tex.z-dn.net/?f=%5Coverrightarrow%20%7BPR%7D%3D%3C-2-0%2C2-0%2C0-1%3E%3D%3C-2%2C2%2C-1%3E)
The nonzero vector orthogonal to the plane through the points P,Q, and R is
![\overrightarrow n=\overrightarrow {PQ}\times \overrightarrow {PR}](https://tex.z-dn.net/?f=%5Coverrightarrow%20n%3D%5Coverrightarrow%20%7BPQ%7D%5Ctimes%20%5Coverrightarrow%20%7BPR%7D)
![\overrightarrow n=\det \begin{pmatrix}i&j&k\\ \:\:\:\:\:-2&3&3\\ \:\:\:\:\:-2&2&-1\end{pmatrix}](https://tex.z-dn.net/?f=%5Coverrightarrow%20n%3D%5Cdet%20%5Cbegin%7Bpmatrix%7Di%26j%26k%5C%5C%20%5C%3A%5C%3A%5C%3A%5C%3A%5C%3A-2%263%263%5C%5C%20%5C%3A%5C%3A%5C%3A%5C%3A%5C%3A-2%262%26-1%5Cend%7Bpmatrix%7D)
Expand along row 1.
![\overrightarrow n=i\det \begin{pmatrix}3&3\\ 2&-1\end{pmatrix}-j\det \begin{pmatrix}-2&3\\ -2&-1\end{pmatrix}+k\det \begin{pmatrix}-2&3\\ -2&2\end{pmatrix}](https://tex.z-dn.net/?f=%5Coverrightarrow%20n%3Di%5Cdet%20%5Cbegin%7Bpmatrix%7D3%263%5C%5C%202%26-1%5Cend%7Bpmatrix%7D-j%5Cdet%20%5Cbegin%7Bpmatrix%7D-2%263%5C%5C%20-2%26-1%5Cend%7Bpmatrix%7D%2Bk%5Cdet%20%5Cbegin%7Bpmatrix%7D-2%263%5C%5C%20-2%262%5Cend%7Bpmatrix%7D)
![\overrightarrow n=i(-9)-j(8)+k(2)](https://tex.z-dn.net/?f=%5Coverrightarrow%20n%3Di%28-9%29-j%288%29%2Bk%282%29)
![\overrightarrow n=-9i-8j+2k](https://tex.z-dn.net/?f=%5Coverrightarrow%20n%3D-9i-8j%2B2k)
![\overrightarrow n=](https://tex.z-dn.net/?f=%5Coverrightarrow%20n%3D%3C-9%2C-8%2C2%3E)
Therefore, the nonzero vector orthogonal to the plane is <-9,-8,2>.