If $$x = a$$, $$y = b$$, $$z = c$$ is a solution of the system of linear equations
$$x + 8y + 7z = 0$$
$$9x + 2y + 3z = 0$$
$$x + y + z = 0$$
Such that the point $$(a, b, c)$$ lies on the plane $$x + 2y + z = 6$$, then $$2a + b + c$$ equals:

Let the required point be $$(a,\,b,\,c)$$. Because it is given that $$(a,\,b,\,c)$$ satisfies the first three homogeneous equations, we begin by solving that system

$$\begin{aligned} x + 8y + 7z &= 0, \\ 9x + 2y + 3z &= 0, \\ x + y + z &= 0. \end{aligned}$$

The third relation is the simplest, so we keep it as a backbone:

$$x + y + z = 0 \quad\Longrightarrow\quad x = -\,y - z.$$

To use only two variables, every occurrence of $$x$$ in the other equations can be replaced by $$-y - z$$. Substituting in the first equation,

$$(-y - z) + 8y + 7z = 0.$$

Simplifying term by term,

$$-y + 8y = 7y,\qquad -z + 7z = 6z,$$

so we obtain

$$7y + 6z = 0.$$

Next, substitute $$x = -y - z$$ into the second equation:

$$9(-y - z) + 2y + 3z = 0.$$

Distribute the $$9$$ inside the parentheses:

$$-9y - 9z + 2y + 3z = 0.$$

Combine the like terms carefully:

$$(-9y + 2y)\;+\;(-9z + 3z) = -7y\;-\;6z = 0.$$

Notice that this is merely the negative of the earlier relation $$7y + 6z = 0$$, so no new information is added; the system is dependent, giving infinitely many solutions. We therefore treat one variable as a parameter.

Choose $$z$$ to be the parameter. Write $$z = t$$, where $$t$$ can be any real number. From $$7y + 6z=0$$ we have

$$7y + 6t = 0 \quad\Longrightarrow\quad y = -\frac{6t}{7}.$$

Using $$x = -y - z$$, substitute the expressions for $$y$$ and $$z$$:

$$x = -\left(-\frac{6t}{7}\right) - t = \frac{6t}{7} - t.$$

Express the subtraction with a common denominator of $$7$$:

$$\frac{6t}{7} - t \;=\; \frac{6t}{7} - \frac{7t}{7} \;=\; -\frac{t}{7}.$$

Hence the general solution of the homogeneous system is

$$x = -\frac{t}{7},\quad y = -\frac{6t}{7},\quad z = t.$$

This family of points is written more compactly as

$$(x,\,y,\,z) = t\!\left(-\frac{1}{7},\, -\frac{6}{7},\,1\right).$$

The given point $$(a,\,b,\,c)$$ must lie not only on this line but also on the separate plane

$$x + 2y + z = 6.$$

Substituting the parametric forms into the plane’s equation, we get

$$\left(-\frac{t}{7}\right) + 2\!\left(-\frac{6t}{7}\right) + t = 6.$$

Compute each term step by step. First, double the second fraction:

$$2\!\left(-\frac{6t}{7}\right) = -\frac{12t}{7}.$$

Now write the entire left‐hand side over the common denominator $$7$$:

$$-\frac{t}{7}\;-\;\frac{12t}{7}\;+\;t = -\frac{t}{7} - \frac{12t}{7} + \frac{7t}{7}.$$

Combine the three numerators:

$$-\frac{t + 12t - 7t}{7} = -\frac{6t}{7}.$$

Thus we arrive at

$$-\frac{6t}{7} = 6.$$

Multiply both sides by $$7$$ to clear the denominator:

$$-6t = 42.$$

Divide by $$-6$$ to isolate $$t$$:

$$t = -7.$$

Insert this value of $$t$$ back into the formulas for $$x,\,y,\,z$$:

$$\begin{aligned} a = x &= -\frac{t}{7} = -\frac{-7}{7} = 1,\\[4pt] b = y &= -\frac{6t}{7} = -\frac{6(-7)}{7} = 6,\\[4pt] c = z &= t = -7. \end{aligned}$$

The required expression is $$2a + b + c$$. Substitute $$a = 1,\; b = 6,\; c = -7$$:

$$2a + b + c = 2(1) + 6 + (-7) = 2 + 6 - 7 = 1.$$

Hence, the correct answer is Option C.