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?
Since the tag is "linear algebra":  If the 3 vectors lie in the same plane, then they are linearly dependent.  So the $3\times 3$ matrix whose rows are the vectors needs to have determinant $0$.  If you set up the determinant, you can subtract the top row from the other two rows and the determinant is equal to $a^2-4a-24=0$ which has solutions $a=2\pm 2\sqrt{7}.$
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:

$$\begin{cases}&amp;I&amp;(1,2,3)(A,B,C)=0\implies &amp;A+2B+3C=0\\{}\\
&amp;II&amp;(1,a+2,6)(A,B,C)=0\implies&amp;A+(a+2)B+6C=0\\{}\\
&amp;III&amp;(1,10,a-1)(A,B,C)=0\implies&amp;A+10B+(a-1)C=0\end{cases}$$

Observe that, for example:

$$\begin{cases}
&amp;II-I :\;\;&amp;aB+3C=0\\{}\\
&amp;III-II:\;\;&amp;(a+8)B+(a-7)C=0\end{cases}$$

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?)