The planes defining the line have normal vectors ![{\displaystyle {\vec {v}}=\left[{\begin{array}{c}1\\2\\3\end{array}}\right]{\text{ and }}{\vec {w}}=\left[{\begin{array}{c}1\\1\\1\end{array}}\right]}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/4f86485a66b300d19698e470ab7746fe25e851a0)
. Their cross-product yields a vector that is parallel to 
:
![{\displaystyle {\vec {v}}\times {\vec {w}}=\left[{\begin{array}{c}-1\\2\\-1\end{array}}\right].}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/8c46d3c26cee2f7829207e658ff37b968fdee998)
For this latter vector to be parallel to the plane given by 
, it must be orthogonal to the vector that is normal to the plane, namely ![{\displaystyle \left[{\begin{array}{c}2\\0\\c\end{array}}\right].}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/f2a5d6f47dba77271cbbdae9c6b6a082c8e09d85)
In other words, we have
![{\displaystyle \left[{\begin{array}{c}-1\\2\\-1\end{array}}\right]\cdot \left[{\begin{array}{c}2\\0\\c\end{array}}\right]=-1\cdot 2+2\cdot 0+-1\cdot c=0,}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/b70f52a8b727d460a60d1e0decab196f4be53ed8)
for which the solution is 
