Let $f\left(x\right)=a{x}^3+b{x}^2+cx+5$  touches the x-axis at $P\left(-2,0\right)$  and cut the y-axis at point $Q$  where its gradient is 3 then $\left(a,b,c\right)=$

$0=a{\left(-2\right)}^3+b{\left(-2\right)}^2+c\left(-2\right)+5$

$0=-8a+4b-2c+5\cdots \cdots \left(1\right)$

Since the curve cuts Y-axis is $x=0$

$y=5$

$\therefore Q\left(0,5\right)$

$y=a{x}^3+b{x}^2+cx+5$

$\left(\frac{dy}{dx}\right)=3a{x}^2+2bx+c$

$m={\left(\frac{dy}{dx}\right)}_{\left(0,5\right)}=c=3$

$\left(1\right)\Rightarrow -8a+4b-2\left(3\right)+5=0$

$-8a+4b=1$

$-8\left(-\frac{1}{2}\right)+4\left(\frac{-3}{9}\right)=1$

$4-3=1$

$1=1$

$\therefore \left(a,b,c\right)=\left(-\frac{1}{2},\frac{-3}{4},3\right)$