In order to find a normal vector to the plane containing these three points, we need to find two vectors spanning this plane. This can be achieved by subtracting the points:
![{\displaystyle {\begin{aligned}(4,1,2)-(3,0,1)&=(1,1,1)\\(4,-2,4)-(4,1,2)&=(0,-3,2)\end{aligned}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/eddadeda74e6f28d6a96dd0737067c8cccfa9383)
To find a normal vector to the plane, we need to take the cross product of these vectors.
![{\displaystyle \left|{\begin{array}{ccc}{\vec {i}}&{\vec {j}}&{\vec {k}}\\1&1&1\\0&-3&2\end{array}}\right|}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/c8013ca7336fea11aabf0170a05ff464a0c6b598)
.
Answer: