If $3{\mathrm{x}}^5-7{\mathrm{x}}^4+4{\mathrm{x}}^3+{\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}=0$ is a reciprocal equation of second type then $(\mathrm{a}, \mathrm{b}, \mathrm{c})=$

$\begin{array}{l}If3x^5-7x^4+4x^3+a{x}^2+bx+c=0is\\ \mathrm{Re}ciprocalEquationof\sec ondtypethen\\ ai=-a_n-ifori=1,2--5\\ \therefore 3=-c,-7=-b,4=-a\\ \Rightarrow a=-4,b=7,c=-3\end{array}$