If $\alpha +\beta +\gamma =0$, then find the value of $\frac{3{\beta }^2+{\alpha }^2+{\gamma }^2}{2{\beta }^2-\alpha \gamma }$.

Given condition:
$$\alpha +\beta +\gamma =0$$

We can rewrite this as:

$$\alpha +\gamma =-\beta$$

Squaring both sides of the equation:

$$(\alpha +\gamma {)}^2=(-\beta {)}^2$$

$${\alpha }^2+{\gamma }^2+2\alpha \gamma ={\beta }^2$$

$${\alpha }^2+{\gamma }^2={\beta }^2-2\alpha \gamma$$

Now, substitute this expression for ${\alpha }^2+{\gamma }^2$ into the numerator of the given fraction:

$$\text{Numerator}=3{\beta }^2+({\alpha }^2+{\gamma }^2)$$

$$\text{Numerator}=3{\beta }^2+({\beta }^2-2\alpha \gamma )$$

$$\text{Numerator}=4{\beta }^2-2\alpha \gamma$$

$$\text{Numerator}=2(2{\beta }^2-\alpha \gamma )$$

Now, substitute back into the full fraction expression:

$$\frac{\text{Numerator}}{\text{Denominator}}=\frac{2(2{\beta }^2-\alpha \gamma )}{2{\beta }^2-\alpha \gamma }$$

Canceling out the common term $(2{\beta }^2-\alpha \gamma )$ from both the numerator and the denominator:

$$\frac{2(2{\beta }^2-\alpha \gamma )}{2{\beta }^2-\alpha \gamma }=2$$

Therefore, the value of the expression is $2$.