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?

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

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