If the system of linear equations $$x - 4y + 7z = g$$; $$3y - 5z = h$$; $$-2x + 5y - 9z = k$$ is consistent, then:

We have the system

$$x - 4y + 7z = g \qquad (1)$$

$$3y - 5z = h \qquad (2)$$

$$-2x + 5y - 9z = k \qquad (3)$$

For the system to be consistent, every solution of the first two equations must also satisfy the third. Therefore we first solve (1) and (2) for $$x$$ and $$y$$ in terms of a free variable (which we choose as $$z$$) and the constants $$g,h$$.

From equation (2) we isolate $$y$$:

$$3y - 5z = h$$

$$\Rightarrow\; 3y = h + 5z$$

$$\Rightarrow\; y = \dfrac{h + 5z}{3} \qquad (4)$$

Next, from equation (1) we isolate $$x$$:

$$x - 4y + 7z = g$$

$$\Rightarrow\; x = g + 4y - 7z$$

Substituting the value of $$y$$ from (4):

$$x = g + 4\!\left(\dfrac{h + 5z}{3}\right) - 7z$$

Bring everything over a common denominator of $$3$$ (so that every term contains the same denominator):

$$x = g + \dfrac{4h + 20z}{3} - \dfrac{21z}{3}$$

Combine the $$z$$ terms in the numerator:

$$x = g + \dfrac{4h + 20z - 21z}{3}$$

$$\Rightarrow\; x = g + \dfrac{4h - z}{3} \qquad (5)$$

Now we substitute the expressions (4) and (5) for $$y$$ and $$x$$ into equation (3). If the system is consistent, that substitution must satisfy (3) for every choice of the free variable $$z$$.

Equation (3) is

$$-2x + 5y - 9z = k$$

Insert (5) for $$x$$ and (4) for $$y$$:

$$-2\!\left[g + \dfrac{4h - z}{3}\right] + 5\!\left(\dfrac{h + 5z}{3}\right) - 9z = k$$

Let us open the brackets step by step. First, the term $$-2x$$:

$$-2x = -2g - 2\!\left(\dfrac{4h - z}{3}\right) = -2g - \dfrac{8h - 2z}{3}$$

Second, the term $$5y$$:

$$5y = 5\!\left(\dfrac{h + 5z}{3}\right) = \dfrac{5h + 25z}{3}$$

Now we write the entire left‐hand side of (3) over the common denominator $$3$$. We keep the integer term $$-2g$$ separate for clarity:

Left side $$= -2g + \dfrac{-\,\!(8h - 2z) + (5h + 25z)}{3} - 9z$$

Convert the final term $$-9z$$ to the same denominator (multiply numerator and denominator by $$3$$):

$$-9z = -\dfrac{27z}{3}$$

So the complete numerator becomes

$$[-(8h - 2z) + (5h + 25z) - 27z]$$

Open the brackets carefully:

$$-(8h - 2z) = -8h + 2z$$

Add $$5h + 25z$$:

$$-8h + 2z + 5h + 25z = -3h + 27z$$

Finally, subtract $$27z$$:

$$-3h + 27z - 27z = -3h$$

Thus the entire fraction reduces to

$$\dfrac{-3h}{3} = -h$$

Putting this together with the integer term $$-2g$$, the left side of equation (3) simplifies completely to

$$-2g - h$$

For consistency we equate this to the right side of (3):

$$-2g - h = k$$

Move every term to one side to form a single relation among $$g,h,k$$:

$$-2g - h - k = 0$$

Multiply through by $$-1$$ to write it in a more familiar positive form:

$$2g + h + k = 0$$

This is exactly the condition listed in Option C.

Hence, the correct answer is Option C.