Answer
$\langle \frac{7\sqrt 6}{18}, \frac{\sqrt 6}{9}, -\frac{\sqrt 6}{18} \rangle$ and $\langle -\frac{7\sqrt 6}{18}, -\frac{\sqrt 6}{9}, \frac{\sqrt 6}{18} \rangle$
Work Step by Step
Step 1. Identify the given vectors $\vec u=0i+j+2k$ and $\vec v=i-2j+3k$
Step 2. We know that the cross product of these two vectors will be perpendicular to both, so we have
$\vec u\times\vec v=(1\times3-2(-2))i+(2\times1-0\times3)j+(0(-2)-1\times1)k==7i+2j-k$
Step 3. Find the unit vector of the result from step 2: $\vec n=\frac{1}{\sqrt {(7)^2+(2)^2+(-1)^2}}(7i+2j-k)
=\frac{\sqrt 6}{18}(7i+2j-k)=\frac{7\sqrt 6}{18}i+\frac{\sqrt 6}{9}j-\frac{\sqrt 6}{18}k
=\langle \frac{7\sqrt 6}{18}, \frac{\sqrt 6}{9}, -\frac{\sqrt 6}{18} \rangle$
Step 4. The second normal vector can be the vector opposite to $\vec n$, so that $\vec {-n}=\langle -\frac{7\sqrt 6}{18}, -\frac{\sqrt 6}{9}, \frac{\sqrt 6}{18} \rangle$
