Find the coordinate vectors of vectors $~V~$ relative to $~B'~$. So we have bases 
$$B = \{(-3,0,-3), (-3,2,1) , (1,6,1)\} $$
and 
$$B' = \{(-6, -6, 0), (-2, -6, 4), (-2, -3, 7)\}$$ of $~\mathbb R^3~$
How to find the coordinate vectors of vectors $~V~$ relative to $~B'~$ if 
$$[v]_b = (1,1,1)$$
Not sure how to solve this problem.
 A: As explained in Azif00's comment, $[v]_B = (1,1,1)$ means, with the basis $B = \{B_1, B_2, B_3\}$, that
\begin{align}
v & = 1 \times B_1 + 1 \times B_2 + 1 \times B_3 \\
& = 1 \times (-3,0,-3) + 1 \times (-3,2,1) + 1 \times (1,6,1) \\
& = (-5,8, -1) \tag{1}\label{eq1}
\end{align}
Finding $v$ relative to $B'$ means finding $[v]_{B'} = (a,b,c)$ where, with the basis $B\,' = \{B\,'_1, B\,'_2, B\,'_3\}$, that
\begin{align}
v & = a \times B\,'_1 + b \times B\,'_2 + c \times B\,'_3 \\
& = a \times (-6,-6,0) + b \times (-2,-6,4) + c \times (-2,-3,7) \\
& = (-6a - 2b - 2c, -6a -6b - 3c, 4b + 7c) \tag{2}\label{eq2}
\end{align}
Comparing the co-ordinates between \eqref{eq1} and \eqref{eq2} gives the following $3$ linear equations of
$$-6a - 2b - 2c = -5 \tag{3}\label{eq3}$$
$$-6a - 6b - 3c = 8 \tag{4}\label{eq4}$$
$$4b + 7c = -1 \tag{5}\label{eq5}$$
Solving these sets of equations for $a,b,c$ will then give you the co-ordinates of $[v]_{B'}$. I trust you can finish the rest yourself.
A: I'm Answering this question for any transformation from Basis $E$ to basis $B$. Let $E = (e_1, e_2, e_3)$ be the old coordinate basis, then any vector $x=(x_1, x_2, x_3)$ can be written as $x = x_1e_1+x_2e_2+x_3e_3$. Let $B = (b_1, b_2, b_3)$ be the new coordinate system. Let $\alpha_1, \alpha_2, \alpha_3$ be the coordinates of $x$ w.r.to the new basis. Then $x$ can be written as $x = \alpha_1b_1+\alpha_2b_2+\alpha_3b_3$. So 
\begin{equation}
\begin{split}
\alpha_1b_1+\alpha_2b_2+\alpha_3b_3 &= x_1e_1+x_2e_2+x_3e_3\\
[b_1, b_2, b_3](\alpha_1, \alpha_2, \alpha_3)^{T} &= [e_1, e_2, e_3] (x_1, x_2, x_3)^{T}\\
(\alpha_1, \alpha_2, \alpha_3)^{T} &= [b_1, b_2, b_3]^{-1} [e_1, e_2, e_3] (x_1, x_2, x_3)^{T}
\end{split}
\end{equation}
Therefore the transformation matrix (from basis $E$ to $B$) is given by $T = [b_1, b_2, b_3]^{-1} [e_1, e_2, e_3]$
