A problem on triangle and it's perpendicular bisectors.

A problem on triangle and it's perpendicular bisectors.

I'm trying to solve the following problem :
"In ¢ABC, coordinates of $B$ are $(−3, 3)$. Equation of the
perpendicular bisector of side $AB$ is $2x + y − 7 = 0$. Equation of
the perpendicular bisector of side $BC$ is $3x − y − 3 = 0$.
Mid point of side $AC$ is $E(11/2,7/2)$. Find $AC^2$."
Here is what I did :
By solving $3x − y − 3 = 0$ and $2x + y − 7 = 0$ I find
that the intersection of perpendicular bisectors is at $(2,3)$ .
Then using the two points $(2,3)$ and $(11/2,7/2)$, I get the equation of
perpendicular bisector of $AC$ as $y = x/7+19/7$.
So the slope of AC is -7 and then using point slope form ,
$y-7/2=-7(x-11/2)$ Thus the equation of line $AC$ is $y = 42-7 x$ .
Similarly equation of line $BC$ is $y = 2-x/3$ .
So $AC$ and $BC$ intersect at $(6,0)$.
By using the fact that $E$ is the midpoint of $AC$, I find Co-ordinates of
A as $(5,7)$.
So the distance between A and C is $5 \sqrt2$, and $AC^2=50$.
But this answer is wrong and the correct answer is $74$ ( I checked the
answer sheet) .
What have I done wrong ?