Change of basis matrix to convert standard basis to another basis
Denote $E$ the canonical basis of $\mathbb{R}^3$.
A) These three column vectors define a $3\times 3$ matrix $$P=\left(\matrix{-1&-1&1\\1&0&1\\0&1&1}\right)$$ which is the matrix of the linear map $$ Id:(\mathbb{R}^3,B)\longrightarrow (\mathbb{R}^3,E). $$ This means in particular that whenever you right multiply it by a column vector $(x_1,x_2,x_3)$ where $x_j$ are the coordinates of a vector $x=x_1B_1+x_2B_2+x_3B_3$ with the respect to the basis $B$, you obtain the coordinates of $x$ in the canonical basis $E$.
What you want is the matrix of $$ Id:(\mathbb{R}^3,E)\longrightarrow (\mathbb{R}^3,B). $$ That is $P^{-1}$, the inverse of the matrix above. This will transform, by right multiplication, the coordinates of a vector with respect to $E$ into its coordinates with respect to $B$. That's the change of basis matrix you need.
B) As explained above, you just have to right multiply the change of basis matrix $P^{-1}$ by this column vector.
Check your answer: you should find
$$P^{-1}=\left(\matrix{-1/3&2/3&-1/3\\-1/3&-1/3&2/3\\1/3&1/3&1/3} \right)$$ $$\left(\matrix{-1/3&2/3&-1/3\\-1/3&-1/3&2/3\\1/3&1/3&1/3} \right)\left(\matrix{1\\0\\0}\right)=\left(\matrix{-1/3\\-1/3\\1/3}\right).$$
By definition change of base matrix contains the coordinates of the new base in respect to old base as it's columns. So by definition $B$ is the change of base matrix. Key to solution is equation $v = Bv'$ where $v$ has coordinates in old basis and $v'$ has coordinates in the new basis (new basis is B-s cols) suppose we know that in old basis $v$ has coords $(1,0,0)$ (as a column) (which is by the way just an old base vector) and we want to know $v'$ (the old base vector coordinates in terms of new base) then from the above equation we get $$B^{-1}v = B^{-1}Bv' \Rightarrow B^{-1}v = v'$$
As a side-node, sometimes we want to ask how does that change of base matrix B act if we look at it as linear transformation, that is given vector v in old base $v=(v_1,...,v_n)$, what is the vector $Bv$? In general it is a vector whith i-th coordinate bi1*v1+...+bin*vn (dot product of i-th row of $B$ with $v$). But in particular if we consider v to be an old base vector having coordinates (0...1...0) (coordinates in respect the old base) where 1 is in the j-th position, then we get $Bv = (b_{1j},...,b_{nj})$ which is the j-th column of B, which is the j-th base vector of the new base. Thus we may say that B viewed as linear transformation takes old base to new base.