y=ax2+bx+c, y'=2ax+b is the gradient. At (1,1) the gradient (tangent) is 2a+b. If the normal at (1,1) is perpendicular to the given line, then the given line is parallel to the tangent. The gradient of 2x+y-3=0 is -2 (y=3-2x) so 2a+b=-2.
Also, (1,1) lies on the parabola so 1=a+b+c.
At x=3 the gradient=0=6a+b, so b=-6a. 2a+b=-2, therefore 2a-6a=-2, -4a=-2, a=½, so b=-3 and:
½-3+c=1, c=3½.
y=½x2-3x+3½, that is, y=x2/2-3x+7/2.
(The given line is in fact the tangent line at (1,1).)