A point on the $X-$ axis which is equidistant from the points  $\left(7,6\right)$  and $\left(3,4\right)$  is

Let $\left(a,0\right)$  be the point on the x-axis that is equidistant from points $\left(7,6\right)$  and $\left(3,4\right)$

$\sqrt{{\left(7-a\right)}^2+{\left(6+0\right)}^2}=\sqrt{{\left(3-a\right)}^2+{\left(4-0\right)}^2}$

$\Rightarrow \sqrt{49+a^2-14a+36}=\sqrt{9+a^2-6a+16}$

$\Rightarrow a^2-14a+85=a^2-6a+25$

$=-14a+6a=25-85$

$-8a=-60$

$a=\frac{60}{8}=\frac{15}{2}$

$\therefore$  The point on x-axis $=\left(\frac{15}{2},0\right)$