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?

Say the plane is $\;Ax+By+Cz+D=0\;$ . Any vector parallel to this plane must be perpendicular to the plane's normal, which is $\;(A,B,C)\;$ , so then it must be:
$$\begin{cases}&I&(1,2,3)(A,B,C)=0\implies &A+2B+3C=0\\{}\\
&II&(1,a+2,6)(A,B,C)=0\implies&A+(a+2)B+6C=0\\{}\\
&III&(1,10,a-1)(A,B,C)=0\implies&A+10B+(a-1)C=0\end{cases}$$
Observe that, for example:
$$\begin{cases}
&II-I :\;\;&aB+3C=0\\{}\\
&III-II:\;\;&(a+8)B+(a-7)C=0\end{cases}$$
Solving this linear system, you get:
$$(II-I)\;\;C=-\frac a3B\stackrel{III-II}\implies (a+8)B+(a-7)\left(-\frac a3B\right)=0\stackrel{\text{assume}\; B\neq0}\implies$$
$$a+8-\frac{a^2}3+\frac73a=0\implies a^2-10a-24=0\iff(a-12)(a+2)=0$$
Take it from here (what if $\;B=0\;$ , say?)
