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?
 A: 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 $a=2\pm 2\sqrt{7}.$
A: 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 $\;a=\color{red}{2(1\pm\sqrt 7)}$.
A: 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?)
