The following system of linear equations
$$7x + 6y - 2z = 0$$
$$3x + 4y + 2z = 0$$
$$x - 2y - 6z = 0$$, has:

We are given the three homogeneous linear equations

$$7x + 6y - 2z = 0$$

$$3x + 4y + 2z = 0$$

$$x - 2y - 6z = 0.$$

Because every constant term on the right-hand side is zero, the origin $$(0,0,0)$$ is always a solution. We must decide whether this is the only solution or whether infinitely many non-trivial solutions also exist.

To do so we use the usual elimination method. First, from the third equation we isolate $$x$$. The rule we now employ is “to solve for one variable, move every other term to the opposite side and then divide by the coefficient of the chosen variable.” Here the coefficient of $$x$$ is $$1$$, so division is unnecessary. We obtain

$$x - 2y - 6z = 0 \;\Longrightarrow\; x = 2y + 6z.$$

Next, we substitute this expression for $$x$$ into the first and second equations so that only $$y$$ and $$z$$ remain. Substituting into the first equation, we write

$$7x + 6y - 2z = 0.$$

Replacing $$x$$ by $$2y + 6z$$ gives

$$7(2y + 6z) + 6y - 2z = 0.$$

Now we expand the bracket according to the distributive law $$a(b+c)=ab+ac$$:

$$14y + 42z + 6y - 2z = 0.$$

Collecting like terms, first for $$y$$ and then for $$z$$, we have

$$\bigl(14y + 6y\bigr) + \bigl(42z - 2z\bigr) = 0,$$

so

$$20y + 40z = 0.$$

Every term now contains the common factor $$20$$, so we divide by $$20$$ (division property of equality) to simplify:

$$y + 2z = 0.$$

Thus

$$y = -2z.$$

We now perform the same substitution in the second original equation

$$3x + 4y + 2z = 0.$$

Replacing $$x$$ again by $$2y + 6z$$ gives

$$3(2y + 6z) + 4y + 2z = 0.$$

Distributing the $$3$$ we obtain

$$6y + 18z + 4y + 2z = 0.$$

Grouping like terms yields

$$\bigl(6y + 4y\bigr) + \bigl(18z + 2z\bigr) = 0,$$

so

$$10y + 20z = 0.$$

Again every term contains the factor $$10$$, and dividing by $$10$$ we get

$$y + 2z = 0.$$

This is precisely the same linear relation between $$y$$ and $$z$$ that we obtained from the first equation, so there is no contradiction. Hence the two substituted equations are consistent with each other and both reduce to the single condition

$$y = -2z.$$

We now return to the expression already found for $$x$$, namely $$x = 2y + 6z$$. Substituting $$y = -2z$$ into that expression, we obtain

$$x = 2(-2z) + 6z = -4z + 6z = 2z.$$

Thus every solution of the system must satisfy simultaneously

$$x = 2z, \qquad y = -2z,$$

with $$z$$ free to take any real value. We therefore have one free parameter—call it $$z = t$$—and we may write the whole solution set compactly as

$$(x, y, z) = (2t,\,-2t,\,t), \qquad t \in \mathbb{R}.$$

Because $$t$$ can be chosen arbitrarily, there are infinitely many solutions, all obeying the linear relation $$x = 2z.$$ The option that matches this description is Option C.

Hence, the correct answer is Option C.