1. Given data

• Point $A(4,3)$
• Point $B$ lies on the line $x-y-2=0 \implies y=x-2$
• Length $AB=\frac{\sqrt{29}}{3}$
• The line through $A$ and $B$ has slope greater than $1$.

We need to find the coordinates of $B$, then check which option it satisfies.

2. Let coordinates of $B$ be

Since $B$ lies on $y=x-2$, let $B=(t,t-2)$

3. Use distance formula

Distance between $A(4,3)$ and $B(t,t-2)$ is $AB^2=(t-4)^2+((t-2)-3)^2$ $=(t-4)^2+(t-5)^2$

Given $AB=\frac{\sqrt{29}}{3}$ so $AB^2=\frac{29}{9}$

Hence, $(t-4)^2+(t-5)^2=\frac{29}{9}$

Expand: $t^2-8t+16+t^2-10t+25=\frac{29}{9}$ $2t^2-18t+41=\frac{29}{9}$

Multiply by $9$: $18t^2-162t+369=29$ $18t^2-162t+340=0$ $9t^2-81t+170=0$

Solve: $t=\frac{81\pm\sqrt{81^2-4\cdot 9\cdot 170}}{18}$ $=\frac{81\pm\sqrt{6561-6120}}{18}$ $=\frac{81\pm 21}{18}$

So, $t=\frac{102}{18}=\frac{17}{3} \quad \text{or} \quad t=\frac{60}{18}=\frac{10}{3}$

Thus possible points are: $B_1=\left(\frac{17}{3},\frac{11}{3}\right), \qquad B_2=\left(\frac{10}{3},\frac{4}{3}\right)$

4. Use the slope condition

Slope of line through $A(4,3)$ and $B$ is $m=\frac{y_B-3}{x_B-4}$

For $B_1=\left(\frac{17}{3},\frac{11}{3}\right)$:

m=\frac{\frac{11}{3}-3}{\frac{17}{3}-4}=rac{\frac{2}{3}}{\frac{5}{3}}=\frac{2}{5}<1 Not allowed.

For $B_2=\left(\frac{10}{3},\frac{4}{3}\right)$:

m=\frac{\frac{4}{3}-3}{\frac{10}{3}-4}=rac{-\frac{5}{3}}{-\frac{2}{3}}=\frac{5}{2}>1 Allowed.

So, $B=\left(\frac{10}{3},\frac{4}{3}\right)$

5. Check the options

Option A: $2x+y=9$

$2\cdot \frac{10}{3}+\frac{4}{3}=\frac{24}{3}=8\neq 9$ False.

Option B: $3x-2y=7$

$3\cdot \frac{10}{3}-2\cdot \frac{4}{3}=10-\frac{8}{3}=\frac{22}{3}\neq 7$ False.

Option C: $x+2y=6$

$\frac{10}{3}+2\cdot \frac{4}{3}=\frac{10}{3}+\frac{8}{3}=\frac{18}{3}=6$ True.

Option D: $2x-3y=3$

$2\cdot \frac{10}{3}-3\cdot \frac{4}{3}=\frac{20}{3}-4=\frac{8}{3}\neq 3$ False.

6. Final answer

The point $B$ lies on $\boxed{x+2y=6}$ So the correct option is C.