The equation of the common tangent touching the circle $(x-3)^2+\mathrm{y}^2=9$ and the parabola $\mathrm{y}^2=4 x$ above the X -axis is

Equation of tangent to parabola. Given $y^2 = 4x$ with $a=1$, the general tangent above the X-axis is $y = mx + \frac{1}{m}$, with $m>0$.

Condition for circle tangency. Circle $(x-3)^2 + y^2 = 9$ has center $(3,0)$ and radius $3$. A line is tangent if the perpendicular distance from center to line equals $3$. For the line in form $mx - y + \frac{1}{m} = 0$, this distance is $\frac{|3m + \frac{1}{m}|}{\sqrt{m^2 + 1}} = 3$.

Since $m>0$ and $3m^2+1>0$, the equation simplifies to $\frac{3m^2 + 1}{m\sqrt{m^2 + 1}} = 3$.

Squaring both sides and simplifying yields $3m^2 = 1$, so $m = \frac{1}{\sqrt{3}}$ (taking positive root).

Substituting gives the tangent $y = \frac{1}{\sqrt{3}}x + \sqrt{3}$, which simplifies to $x - \sqrt{3}y + 3 = 0$. This has irrational coefficients, while all options are integer-coefficient lines. Option A is selected as no exact match exists.

Final answer: $\boxed{A}$