We will use the following approach. Note that the difference between two points on a plane is a vector that is parallel to the plane. If we can find two such vectors (that are not co-linear), then their cross-product is normal to the plane and can be used to write down an equation that the points on the plane need to satisfy.
Let then
![{\displaystyle {\begin{aligned}{\vec {v}}&=(2,-1,1)-(1,1,0)=&\left[{\begin{array}{c}1\\-2\\1\end{array}}\right]\\{\vec {w}}&=(1,1,0)-(0,1,c)=&\left[{\begin{array}{c}1\\0\\-c\end{array}}\right].\end{aligned}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/5568f456e84d333216ef20c25e17d1cbb0a45c7c)
Their cross-product is
![{\displaystyle {\vec {v}}\times {\vec {w}}=\left[{\begin{array}{c}2c\\1+c\\2\end{array}}\right].}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/2a85fbc4f7737f599a191c0923830616ef17a166)
Therefore, a point 
in the plane is characterized by the property that the vector from 
(or any other point on the plane) to is orthogonal to 
:
Substituting 
for yields the equation
for which 
is the only solution.
