$\text{ If the function }g\left(x\right)=\left\{\begin{array}{c}\frac{e^{px}+{\log }_e(1+4x)+q}{x^3},x\ne 0\text{ is continuous at }x=0,\\ r,x=0\end{array}\right.$  $\text{ then the value of }6pqr\text{ is }$

$\text{ Since }g\left(x\right)\text{ is continuous at }x=0\text{ then }$

$\begin{array}{l}\underset{x\to 0}{lim} \frac{e^{px}+{\log }_e(1+4x)+q}{x^3}=r\\ \therefore \underset{x\to 0}{lim} \left\{\frac{\left(1+\left(px\right)+\frac{p^2{x}^2}{2!}+\frac{p^3{x}^3}{3!}+\dots \dots \right)+\left(4x-\frac{16x^2}{2}+\frac{64}{3}{x}^3+\dots \dots \right)+q}{x^3}\right\}=r\end{array}$

$\begin{array}{l}\Rightarrow \underset{x\to 0}{lim} \left\{\frac{(1+q)+(p+4)x+\left(\frac{p^2}{2}-8\right)x^2+\left(\frac{p^3}{6}+\frac{64}{3}\right)x^3+\dots ..}{x^3}\right\}=r\\ \therefore 1+q=0,p+4=0\text{ and }\frac{p^2}{2}-8=0\text{ and }\frac{p^3}{6}+\frac{64}{3}=r\\ \therefore q=-1,p=-4,r=\frac{32}{3}\\ \therefore 6pqr=6(-4)(-1)\left(\frac{32}{3}\right)=256\end{array}$