Affine map $f : P_1 \to P_2$ between two planes I'm learning affine geometry, specifically affine maps, and need help with the following problem :

We give the affine planes
$$P_1 = \{(x, y, z) \in \mathbb R^3 : 3x + 2y + z = 6\} \quad \text{and} \quad P_2 = \{(x, y, z) \in \mathbb R^3 : -x - 2y + 3z = 2\}$$
$(1)$ Find affine bases $(a_0, a_1, a_2)$ and $(a_0', a_1', a_2')$ of $P_1, P_2$ respectively with $a_i$ (resp. $a_i'$) the points of intersection of the planes with the three axes $x, y, z$.
$(2)$ Find the affine transformation $f : P_1 \to P_2$, with respect to the above bases, which send $a_0$ to $-a_0'+2a_2'$, $a_1$ to $(a_0' + a_1' + a_2')/3$ and $a_2$ to $(a_1' + a_2')/2$. Is it regular (invertible)? Find the inverse image of the point $3a_1' - 5a_2'$.

Since I'm having difficulties for $(2)$ I'm going to share my work for $(1)$.
$(1)$ We first determine the desired basis for the plane $P_1$. To find the $x$-intercept we set the $y$ and $z$ coordinates equal to zero. This gives
$$3x + 2(0) + 0 = 6 \iff x = 2 \implies a_1 = (2, 0, 0).$$
We proceed similarly for the $y$ and $z$-intercept. We find the vectors $a_1 = (0, 3, 0)$ and $a_2 = (0, 0, 6)$. Hence
$$(a_0, a_1, a_2) = ((2, 0, 0), (0, 3, 0), (0, 0, 6))$$
is the desired ordered affine basis for $P_1$.
Using the same methodology, the desired affine basis for $P_2$ is given by
$$(a_0', a_1', a_2') = ((-2, 0, 0), (0, -1, 0), (0, 0, 2/3)).$$

Is my work correct for $(1)$? Unfortunately I have no idea how to solve $(2)$. I'm looking for an answer which makes use of the following definition :

Definition. An affine transformation of $\mathbb R^n$ is a function $t : \mathbb R^n \to \mathbb R^n$ of the form $$t(\mathbf x) = \mathbf{Ax} + \mathbf b, $$ where $\mathbf A$ is an invertible $n \times n$ matrix and $\mathbf b \in \mathbb R^n$.

 A: A big clue:You already got an affine basis for $P_1$, say $B=\{a_1 , a_2, a_3\}$.       $f: P_1 \rightarrow P_2$ is an affine function,  therefore, $f$ is completely identified provided that we know the values of $f$ on the basis $B$. Note that for any $a \in P_1$  there are unique $\lambda_1 ,\lambda_2 ,\lambda_3 \in R$  with  $\lambda_1 +\lambda_2 +\lambda_3=1$   such that $a=\lambda_1a_1 +\lambda_2 a_2+\lambda_3 a_3$. Now since $f$ is affine 
$$f(a)=f(\lambda_1a_1 +\lambda_2 a_2+\lambda_3 a_3)=\lambda_1f(a_1) +\lambda_2 f(a_2) +\lambda_3 f(a_3)$$
and you know what $f(a_i)$'s are!
In order to find a closed form of $f(x,y,z)$ where $(x,y,z) \in P_1$ ,  you must determine $\lambda_i$'s in terms of $x,y,z$. 
EDIT:  using the condition $\lambda_1 +\lambda_2 +\lambda_3=1$, you can drop one of $\lambda_i s$ in representation of $f$, say you choose $\lambda_3$. 
Therfore ,
$$f(a)=\lambda_1f(a_1) +\lambda_2 f(a_2) +(1-\lambda_1 - \lambda_2) f(a_3)$$
Let $a= (x,y,z)\in P_1 $ so you have 
$$    \lambda_1a_1 +\lambda_2 a_2+(1-\lambda_1 - \lambda_2) a_3  = (x,y,z) $$
this give you a system of linear equations with variables $\lambda_1$ , $\lambda_2 \in \Bbb R.$
Now your task is to find $\lambda_i s$ using Gausing Elimination, and plugin back to $f$ formula. 
Good luck!
A: Your work for (1) is correct.
Here is my proposal for (2) (I hope it will be useful):
$P_1$ is the affine plane containing $a_0$ and directed by the vector plane generated by $a_1-a_0$ and $a_2-a_0$; $P_2$ is the affine plane containing $a_0'$ and directed by the vector plane generated by $a_1'-a_0'$ and $a_2'-a_0'$. Having chosen $a_0$ as the origin in $P_1$, we can equate each element $x$ of $P_1$ with the pair $(\xi,\eta)$ such that $x-a_0=\xi(a_1-a_0)+\eta(a_2-a_0)$; similarly in $P_2$, where we note such an element $(\xi',\eta')$. Then $(\xi',\eta')=f(x)=A.x+b$. $f(a_1)-f(a_0)=Aa_1+b-(Aa_0+b)=Aa_1-Aa_0=A(a_1-a_0)$ since $A$ is linear. We then obtain $A(a_1-a_0)=\frac13(a_1'-a_0')-\frac53(a_2'-a_0')$. Similarly for $A(a_2-a_0)$. Finally, the matrix of $A$ relative to these bases is $\begin{bmatrix} \frac13 & \frac12 \\ -\frac53 & -\frac32 \end{bmatrix}$ .
To obtain $b$, we write that $f(a_0)=f((0,0))=b=-a_0'+2a_2'=a_0'+2(a_2'-a_0')=(0,2)$. Hence the expression for f required : $$f(x)=\begin{bmatrix} \frac13 & \frac12 \\ -\frac53 & -\frac32 \end{bmatrix}x+\begin{bmatrix} 0 \\ 2 \end{bmatrix}$$
$f$ is regular, since $\begin{vmatrix} \frac13 & \frac12 \\ -\frac53 & -\frac32 \end{vmatrix} \neq 0$.
As for $3a_1' - 5a_2'$, I agree with what @Demophilus and @san have written about it.
