Let the equation of the plane, that passes through the point $$(1, 4, -3)$$ and contains the line of intersection of the planes $$3x - 2y + 4z - 7 = 0$$ and $$x + 5y - 2z + 9 = 0$$, be $$\alpha x + \beta y + \gamma z + 3 = 0$$, then $$\alpha + \beta + \gamma$$ is equal to:

We have two given planes whose equations are $$3x-2y+4z-7=0$$ and $$x+5y-2z+9=0$$. A plane that contains their line of intersection can be written, by the concept of the family of planes, as

$$\bigl(3x-2y+4z-7\bigr)+\lambda\bigl(x+5y-2z+9\bigr)=0,$$ where $$\lambda$$ is a real parameter.

Expanding and collecting the coefficients of $$x$$, $$y$$, $$z$$ and the constant term, we get

$$\bigl(3+\lambda\bigr)x+\bigl(-2+5\lambda\bigr)y+\bigl(4-2\lambda\bigr)z+\bigl(-7+9\lambda\bigr)=0.$$

This required plane must also pass through the point $$(1,4,-3)$$, so we substitute $$x=1,\;y=4,\;z=-3$$ into the above equation:

$$\bigl(3+\lambda\bigr)(1)+\bigl(-2+5\lambda\bigr)(4)+\bigl(4-2\lambda\bigr)(-3)+\bigl(-7+9\lambda\bigr)=0.$$

Simplifying each term separately:

First term: $$3+\lambda.$$

Second term: $$(-2+5\lambda)\cdot4=-8+20\lambda.$$

Third term: $$(4-2\lambda)\cdot(-3)=-12+6\lambda.$$

Fourth term: $$-7+9\lambda.$$

Adding all these terms:

$$\bigl(3+\lambda\bigr)+\bigl(-8+20\lambda\bigr)+\bigl(-12+6\lambda\bigr)+\bigl(-7+9\lambda\bigr)=0.$$

Combining the constant parts $$3-8-12-7=-24$$ and the $$\lambda$$ parts $$\lambda+20\lambda+6\lambda+9\lambda=36\lambda$$, we obtain

$$-24+36\lambda=0.$$

Solving for $$\lambda$$ gives

$$36\lambda=24\quad\Longrightarrow\quad\lambda=\dfrac{24}{36}=\dfrac23.$$

Now we substitute $$\lambda=\dfrac23$$ back into the coefficients of $$x$$, $$y$$ and $$z$$:

$$\alpha=3+\lambda=3+\dfrac23=\dfrac{11}{3},$$

$$\beta=-2+5\lambda=-2+5\left(\dfrac23\right)=-2+\dfrac{10}{3}=\dfrac{4}{3},$$

$$\gamma=4-2\lambda=4-2\left(\dfrac23\right)=4-\dfrac43=\dfrac{8}{3}.$$

The constant term with this value of $$\lambda$$ is

$$-7+9\lambda=-7+9\left(\dfrac23\right)=-7+6=-1.$$

The problem statement wants the plane in the form $$\alpha x+\beta y+\gamma z+3=0$$, i.e., the constant term should be $$+3$$, not $$-1$$. To convert $$-1$$ to $$+3$$ we multiply the entire equation by $$-3$$. Doing so multiplies each coefficient by $$-3$$:

$$\alpha' = -3\left(\dfrac{11}{3}\right)=-11,\qquad \beta' = -3\left(\dfrac{4}{3}\right)=-4,\qquad \gamma' = -3\left(\dfrac{8}{3}\right)=-8.$$

Thus the desired plane is $$-11x-4y-8z+3=0,$$ which matches the required constant term.

Finally, the sum of the coefficients is

$$\alpha'+\beta'+\gamma' = -11-4-8=-23.$$

Hence, the correct answer is Option C.