Solution
= Solution
A direction of the intersection <line> must be <orthogonal> to both <normal vectors>. Their <cross product> is
$$
(2,6,-3)\times(5,2,-1)=(0,-13,-26).
$$
Thus a convenient parallel <vector> is $\boxed{u=(0,1,2)}$.
= Solution
A direction of the intersection <line> must be <orthogonal> to both <normal vectors>. Their <cross product> is
$$
(2,6,-3)\times(5,2,-1)=(0,-13,-26).
$$
Thus a convenient parallel <vector> is $\boxed{u=(0,1,2)}$.