Finding fixed length perpendicular vectors to $\vec{a}$ and $\vec{b}$

I need to find the components of vectors $ \vec{c}_{i}$ that are perpendicular to $ \vec{a} = [3, -1, 5]$ and $ \vec{b} = [2, 2, 5]$ , knowing that the length of obtained vectors $ \vec{c}_{i}$ has to be exactly $ \sqrt{42}$ .

I know two things:

  1. Vector $ \vec{c} = [c_{1}, c_{2}, c_{3}]$ is perpendicular to $ \vec{a} = [a_{1}, a_{2}, a_{3}]$ if their dot product equals zero. Thus the following condition has to be satisfied: $ c_{1}a_{1} + c_{2}a_{2} + c_{3}a_{3} = 0$ .
  2. Cross product of vectors $ \vec{a}$ and $ \vec{b}$ would give me as a result vector $ \vec{c}$ that is indeed perpendicular to both vectors. However, is this information useless for this task?

Anyway how should I solve this task?

Perhaps solving the following system of equations would yield the vectors I am looking for?

$ $ \begin{cases} a_{1}c_{1} + a_{2}c_{2} + a_{3}c_{3} = 0 \ b_{1}c_{1} + b_{2}c_{2} + b_{3}c_{3} = 0 \ \sqrt{c^2_{1} + c^2_{2} + c^2_{3}} = \sqrt{42} \end{cases}$ $

$ $ \begin{cases} 3c_{1} -c_{2} + 5c_{3} = 0 \ 2c_{1} + 2c_{2} + 5c_{3} = 0 \ \sqrt{c^2_{1} + c^2_{2} + c^2_{3}} = \sqrt{42} \end{cases}$ $