Let $$\alpha, \beta$$ be roots of the equation $$x^2 - 3x + r = 0$$, and $$\frac{\alpha}{2}, 2\beta$$ be roots of the equation $$x^2 + 3x + r = 0$$.
If roots of the equation $$x^2 + 6x = m$$ are $$2\alpha + \beta + 2r$$ and $$\alpha - 2\beta - \frac{r}{2}$$, then $$m$$ equals to :

The first quadratic is $$x^2-3x+r=0$$ whose roots are $$\alpha,\,\beta$$.

For a quadratic $$x^2+px+q=0$$ we know
  • sum of roots $$=\,-p$$
  • product of roots $$=\,q$$.

Therefore, from $$x^2-3x+r=0$$ we get
$$\alpha+\beta = 3\quad -(1)$$
$$\alpha\beta = r\quad\; -(2)$$

The second quadratic is $$x^2+3x+r=0$$ whose roots are $$\dfrac{\alpha}{2},\;2\beta$$. Applying the same relations,

Sum of roots:
$$\frac{\alpha}{2}+2\beta = -3\quad -(3)$$

Product of roots:
$$\frac{\alpha}{2}\,(2\beta)=\alpha\beta=r$$ - this is already consistent with $$(2)$$.

We now solve $$(1)$$ and $$(3)$$ for $$\alpha$$ and $$\beta$$.

From $$(1):\; \alpha = 3-\beta.$$ Substitute in $$(3):$$

$$\frac{3-\beta}{2}+2\beta = -3$$ $$\Rightarrow 3-\beta+4\beta = -6$$ $$\Rightarrow 3+3\beta = -6$$ $$\Rightarrow 3\beta = -9$$ $$\Rightarrow \beta = -3.$$

Then $$\alpha = 3-(-3)=6.$$ Using $$(2):\; r=\alpha\beta = 6(-3) = -18.$$

The third quadratic is given as $$x^2+6x = m$$, whose two roots are $$x_1 = 2\alpha+\beta+2r,\qquad x_2 = \alpha-2\beta-\frac{r}{2}.$$

Compute each root with the obtained values:

First root:
$$x_1 = 2(6)+(-3)+2(-18)=12-3-36=-27.$$

Second root:
$$x_2 = 6-2(-3)-\frac{-18}{2}=6+6+9=21.$$

For the quadratic $$x^2+6x=m$$ write it in standard form: $$x^2+6x-m=0.$$ Hence,   • sum of roots $$= -6$$ (matches $$x_1+x_2=-27+21=-6$$),
  • product of roots $$= -m$$.

Product using the calculated roots:
$$x_1x_2 = (-27)(21) = -567 = -m.$$ Therefore, $$m = 567.$$

Option D which is: $$567$$