Linear algebra find a parallel to the same plane Find ''$a$'' for which all three vectors $(1,2,3)$,$(1,2+a,6)$ and $(1,10,a-1)$
are parallel to the same plane.
I am more or less sure about the process of doing this.
Iam thinking about doing dot the product of each vector and solve for $a$.
Any hints on this?

You have to determine at which condition the three vectors are collinear. You can use the determinantal criterion:
$$0=\begin{vmatrix}1&1&1\\2&2+a&10\\3&6&a-1\end{vmatrix}=\begin{vmatrix}1&0&0\\2&a&8\\3&3&a-4\end{vmatrix}=a(a-4)-24=a^2-4a-24=(a-2)^2-28$$
so $\;a=\color{red}{2(1\pm\sqrt 7)}$.
