If $$x^2 - y^2 + 2hxy + 2gx + 2fy + c = 0$$ is the locus of a point, which moves such that it is always equidistant from the lines $$x + 2y + 7 = 0$$ and $$2x - y + 8 = 0$$, then the value of $$g + c + h - f$$ equals

The distance of a point $$P(x,y)$$ from a straight line $$ax+by+c=0$$ is
$$\dfrac{\left|ax+by+c\right|}{\sqrt{a^{2}+b^{2}}}\, .$$

For a point that is equidistant from the two given lines
$$L_1:\;x+2y+7=0$$ and $$L_2:\;2x-y+8=0$$
we must have

$$\dfrac{\left|x+2y+7\right|}{\sqrt{1^{2}+2^{2}}}= \dfrac{\left|2x-y+8\right|}{\sqrt{2^{2}+(-1)^{2}}}\, .$$

Since $$1^{2}+2^{2}=5$$ and $$2^{2}+(-1)^{2}=5$$, the denominators are equal, so squaring both sides eliminates the absolute value and the square roots:

$$(x+2y+7)^2 = (2x - y + 8)^2\, .$$

Expanding the squares:

Left side:
$$(x+2y+7)^2 = x^{2}+4xy+4y^{2}+14x+28y+49.$$

Right side:
$$(2x-y+8)^2 = 4x^{2}-4xy+y^{2}+32x-16y+64.$$

Equating and bringing every term to the left:

$$x^{2}+4xy+4y^{2}+14x+28y+49 -\bigl(4x^{2}-4xy+y^{2}+32x-16y+64\bigr)=0.$$

Simplifying term by term:
$$-3x^{2}+8xy+3y^{2}-18x+44y-15=0.$$

Multiplying by $$-1$$ makes the coefficient of $$x^{2}$$ positive:

$$3x^{2}-8xy-3y^{2}+18x-44y+15=0.$$

Dividing by $$3$$ (so that the coefficient of $$x^{2}$$ becomes $$1$$):

$$x^{2}-\dfrac{8}{3}xy-y^{2}+6x-\dfrac{44}{3}y+5=0.$$

Now compare with the general form
$$x^{2}-y^{2}+2hxy+2gx+2fy+c=0.$$

Thus we read off:
$$2h=-\dfrac{8}{3}\;\Longrightarrow\;h=-\dfrac{4}{3},$$
$$2g=6\;\Longrightarrow\;g=3,$$
$$2f=-\dfrac{44}{3}\;\Longrightarrow\;f=-\dfrac{22}{3},$$
$$c=5.$$

We need $$g+c+h-f$$:

$$g+c+h-f =3+5+\Bigl(-\dfrac{4}{3}\Bigr)-\Bigl(-\dfrac{22}{3}\Bigr) =8+\dfrac{-4+22}{3} =8+\dfrac{18}{3} =8+6 =14.$$

Therefore, $$g+c+h-f=14,$$ which is Option A.