If the roots of the equation $$bx^2 + cx + a = 0$$ be imaginary, then for all real values of $$x$$, the expression $$3b^2x^2 + 6bcx + 2c^2$$ is

The quadratic $$bx^{2}+cx+a=0$$ has imaginary (non-real) roots.
This happens exactly when its discriminant is negative:

$$\Delta = c^{2}-4ab \lt 0 \quad\Longrightarrow\quad 4ab-c^{2} \gt 0$$

Now consider the required expression

$$E(x)=3b^{2}x^{2}+6bcx+2c^{2}$$

Rewrite it by completing the square:

$$\begin{aligned} E(x) &= 3\Bigl(b^{2}x^{2}+2bcx+c^{2}\Bigr)-c^{2} \\ &= 3\,(bx+c)^{2}-c^{2}. \end{aligned}$$

Add and subtract $$4ab$$ so that the positive quantity $$4ab-c^{2}$$ appears:

$$\begin{aligned} E(x) &= 3\,(bx+c)^{2}-c^{2}+4ab-4ab \\ &= 3\,(bx+c)^{2}+\bigl(4ab-c^{2}\bigr)-4ab. \end{aligned}$$

Hence

$$E(x)+4ab = 3\,(bx+c)^{2}+\bigl(4ab-c^{2}\bigr).$$

The first term $$3\,(bx+c)^{2}$$ is $$\ge 0$$ for every real $$x$$, and the second term $$4ab-c^{2}$$ is strictly positive from the discriminant condition. Therefore

$$E(x)+4ab \gt 0 \quad\text{for all real } x.$$

This inequality rearranges to

$$E(x) \gt -4ab \quad\text{for all real } x.$$

Thus the expression $$3b^{2}x^{2}+6bcx+2c^{2}$$ is always greater than $$-4ab$$.

Option C which is: greater than $$-4ab$$