Equation of the ellipe is $\frac{x^2}{6}+\frac{y^2}{3}=1$

The tangent at the point of shortest distance from the line $x+y=7$ should be parallel to the given line

Any point on the given ellipse is $(\sqrt{6}\cos \mathrm{\theta }, \sqrt{3}\sin \mathrm{\theta })$

Equation of the tangent is

$\frac{x\cos \mathrm{\theta }}{\sqrt{6}}+\frac{y\sin \mathrm{\theta }}{\sqrt{3}}=1.$ It is parallel to $x+y=7$

$\because$ Equation of the tangent at $P\left(a\cos \theta ,b\sin \theta \right)$ on ellipse is $\frac{x\cos \theta }{a}+\frac{y\sin \theta }{b}=1$

$\Rightarrow \frac{\cos \mathrm{\theta }}{\sqrt{6}}=\frac{\sin \mathrm{\theta }}{\sqrt{3}}$

$\Rightarrow \frac{\cos \mathrm{\theta }}{\sqrt{2}}=\frac{\sin \mathrm{\theta }}{1}$$\Rightarrow \sin \theta =\frac{1}{\sqrt{3}},\cos \theta =\sqrt{\frac{2}{3}}$

The required point is $(2, 1)$

Hence, $\left(a,b\right)=\left(2,1\right)$

therefore, $\frac{a}{b}=2$.