If $\alpha$ and $\beta$ are the zeroes of the quadratic equation $\left(x²-3x+p\right)=0$ such that $2\alpha +3\beta =15$, then find the value of $p$.

Given that $\alpha$ and $\beta$ are zeroes of the quadratic equation $\left(x²-3x+p\right)=0$ such that $2\alpha +3\beta =15$. 

Determine the sum and product of zeroes. 

Sum of zeroes = $-\frac{b}{a}$

$\Rightarrow \alpha +\beta =-\frac{(-3)}{1}$

$\Rightarrow \alpha +\beta =3$ -----(1)

Product of zeroes = $\frac{c}{a}$

$\Rightarrow \alpha \beta =\frac{p}{1}$

$\Rightarrow \alpha \beta =p$ -----(2) 

Put the value $\alpha =3-\beta$ in $2\alpha +3\beta =15,$ we get,

$2\alpha +3\beta =15$

$\Rightarrow 2\left(3-\beta \right)+3\beta =15$

$\Rightarrow 6-2\beta +3\beta =15$

$\Rightarrow \beta =9$  

From equation (1), we get,

$\alpha =3-\beta$  

$\Rightarrow \alpha =3-9$

$\Rightarrow \alpha =-6$

From equation (2), we get,

$p=\alpha \beta$ 

$\Rightarrow p=-6\times 9$

$\Rightarrow p=-54$

Hence, the value of $p$ is $-54$.