For the differentiable function $f:\mathrm{ℝ}-\left\{0\right\}-\mathrm{ℝ},$ let $3f\left(x\right)+2f\left(\frac{1}{x}\right)=\frac{1}{x}-10,$ then $\left|f\left(3\right)+f^{\text{'}}\left(\frac{1}{4}\right)\right|$ is equal to

Given:

$3f\left(x\right)+2f\left(\frac{1}{x}\right)=\frac{1}{x}-10 ....\left(1\right)$

Replace $x\to \frac{1}{x}$

$3f\left(\frac{1}{x}\right)+2f\left(x\right)=x-10 ....\left(2\right)$

Eliminating $f\left(\frac{1}{x}\right)$ from $\left(1\right) \& \left(2\right)$, we get

$\Rightarrow 5f\left(x\right)=\frac{3}{x}-2x-10$

Let us replace $x$ with $3$

$\Rightarrow 5f\left(3\right)=\frac{3}{3}-2\left(3\right)-10$

$\Rightarrow f\left(3\right)=-3$

$\Rightarrow 5f^{\text{'}}\left(x\right)=\frac{-3}{x^2}-2$

$\Rightarrow 5f^{\text{'}}\left(\frac{1}{4}\right)=-48-2$

$\Rightarrow f^{\text{'}}\left(\frac{1}{4}\right)=-10$

$\Rightarrow \left|f\left(3\right)+f^{\text{'}}\left(\frac{1}{4}\right)\right|=13$

Hence this is the correct option.