If $\alpha$ and $\beta$ are the roots of the equation ${\mathrm{x}}^2+6\mathrm{x}+\mathrm{\lambda }=0$ and $3\mathrm{\alpha }+2\mathrm{\beta }=-20$, then $\mathrm{\lambda }=$

$\mathrm{\alpha }+\mathrm{\beta }=-6 ...\left(\mathrm{i}\right)$

$\mathrm{\alpha \beta }=\mathrm{\lambda } ...\left(\mathrm{ii}\right)$

and given $3\mathrm{\alpha }+2\mathrm{\beta }=-20 ...\left(\mathrm{iii}\right)$

Solving (i) and (iii), we get $\mathrm{\beta }=2, \mathrm{\alpha }=-8$

Substituting these values in (ii), we get $\lambda =-16.$