Is it a vector space? \begin{align*}W=\left\{\left\{x_1,x_2,x_3\right\}|x_1-1=\frac{x_2-2}{2}=\frac{x_3-3}{3}\right\}\end{align*}

\begin{align*}\alpha _1=\left\{x_1,x_2,x_3\right\};\\\\&amp;\\\alpha _2=\left\{y_1,y_2,y_3\right\};\end{align*}

Is $W$ a vector space? I know $\alpha _1+\alpha _2$ and to check whether fit the equation, but seems stuck, and have a little problem in the expression.

$x_1+y_1-2=\frac{x_2+y_2-4}{2}=\frac{x_3+y_3-6}{3}$



subtract 1 from both sides,

$x_1+y_1-1=\frac{x_2+y_2-2}{2}=\frac{x_3+y_3-3}{3}$

Now we are done.
At first, you should add $1$ to the equality constraint describing your set. Then, you get
$$
x_1 = \frac{x_2}2 = \frac{x_3}3
$$
Now, it should be easy to prove that it is a vector space.
It is not Stright line through origin in R3-space, that's why it cant be the subspace and hence vector space as it doesnot contain in this case the zero vector.