Option: A is the correct answer.
The axis of symmetry is:
A. x=3/2
We are given a graph as:
[tex]y+3x-6=-3(x-2)^2+4[/tex]
and
[tex]y=-3(x-2)^2+4+6-3x[/tex]
and
[tex]y=-3[(x-2)^2+x]+10\\\\i.e.\\\\y=-3[x^2+4-4x+x]+10\\\\i.e.\\\\y=-3[x^2-3x+4]+10[/tex]
This equation could also be written in the form:
[tex]y=-3[x^2-3x+4-\dfrac{9}{4}+\dfrac{9}{4}]+10\\\\i.e.\\\\y=-3[(x-\dfrac{3}{2})^2+4-\dfrac{9}{4}]+10\\\\i.e.\\\\y=-3[(x-\dfrac{3}{2})^2+\dfrac{7}{4}]+10\\\\i.e.\\\\y=-3(x-\dfrac{3}{2})^2-3\times \dfrac{7}{4}+10\\\\i.e.\\\\y=-3(x-\dfrac{3}{2})^2+\dfrac{19}{4}[/tex]
Now, we know that for any general equation of the type:
[tex]y=a(x-h)^2+k[/tex]
The axis of symmetry of the graph of the equation is:
[tex]x=h[/tex]
Here we have:
[tex]h=\dfrac{3}{2}[/tex]
Hence, the answer is: Option: A
