Let $\mathrm{f}: \mathbb{R}-\{2\} \rightarrow \mathbb{R}-\{1\}$ defined by $\mathrm{f}(x)=\frac{x-3}{x-2}$ and $\mathrm{g}: \mathbb{R} \rightarrow \mathbb{R}$ defined by $\mathrm{g}(x)=3 \mathrm{x}-2$, then sum of all values of $x$ for which $\mathrm{f}^{-1}(x)+\mathrm{g}^{-1}(x)=\frac{19}{6}$ is

Given functions are $f(x) = \frac{x - 3}{x - 2}$ and $g(x) = 3x - 2$.

The inverse function $f^{-1}(x)$ is determined by setting $y = f(x)$ and solving for $x$ after interchanging variables:

$x = \frac{y - 3}{y - 2} \implies x(y - 2) = y - 3 \implies xy - y = 2x - 3 \implies y(x - 1) = 2x - 3 \implies f^{-1}(x) = \frac{2x - 3}{x - 1}$, with domain $\mathbb{R} - \{1\}$.

Similarly, $g^{-1}(x)$ yields $x = 3y - 2 \implies y = \frac{x + 2}{3}$, so $g^{-1}(x) = \frac{x + 2}{3}$, with domain $\mathbb{R}$.

Set $f^{-1}(x) + g^{-1}(x) = \frac{19}{6}$:

$\frac{2x - 3}{x - 1} + \frac{x + 2}{3} = \frac{19}{6}$.

Combining terms with common denominator $3(x - 1)$:

$\frac{3(2x - 3) + (x - 1)(x + 2)}{3(x - 1)} = \frac{19}{6}$.

Expanding numerator: $6x - 9 + x^2 + 2x - x - 2 = x^2 + 7x - 11$.

$\frac{x^2 + 7x - 11}{3(x - 1)} = \frac{19}{6}$.

Cross-multiplying: $6(x^2 + 7x - 11) = 57(x - 1)$.

$6x^2 + 42x - 66 = 57x - 57 \implies 6x^2 - 15x - 9 = 0$.

Dividing by 3: $2x^2 - 5x - 3 = 0$.

Factoring: $(2x + 1)(x - 3) = 0$, giving roots $x = -\frac{1}{2}$ and $x = 3$.

Both values are valid since $x \neq 1$ is satisfied.

The sum of roots is $3 + (-\frac{1}{2}) = \frac{5}{2}$, which matches option A.