Value of Linear Expression for Point on Plane Satisfying Matrix Equation

To find the value of $7a + b + c$, we start by evaluating the matrix equation given in the passage:

$$\left[\begin{array}{ccc} a & b & c \end{array}\right] \left[\begin{array}{ccc} 1 & 9 & 7 \\ 8 & 2 & 7 \\ 7 & 3 & 7 \end{array}\right] = \left[\begin{array}{ccc} 0 & 0 & 0 \end{array}\right]$$

Carrying out the matrix multiplication gives a homogeneous system of three linear equations:

$$\begin{align} a + 8b + 7c &= 0 \quad &(1) \\ 9a + 2b + 3c &= 0 \quad &(2) \\ 7a + 7b + 7c &= 0 \implies a + b + c = 0 \quad &(3) \end{align}$$

From equation $(3)$, we have:
$c = -a - b$

Substituting $c = -a - b$ into equation $(1)$:
$a + 8b + 7(-a - b) = 0$
$-6a + b = 0 \implies b = 6a$

Now, substituting $b = 6a$ into the expression for $c$:
$c = -a - 6a = -7a$

Let us verify that these relations satisfy equation $(2)$:
$9a + 2(6a) + 3(-7a) = 9a + 12a - 21a = 0$
Equation $(2)$ is identically satisfied. Thus, the coordinates of the point $P(a, b, c)$ can be expressed in terms of a real parameter $k$:
$a = k, \quad b = 6k, \quad c = -7k$

It is given that the point $P(a, b, c)$ lies on the plane:
$2x + y + z = 1$

Substituting the coordinates of $P$:
$2a + b + c = 1$
$2(k) + 6k + (-7k) = 1$
$k = 1$

Therefore, the values of $a, b,$ and $c$ are:
$a = 1, \quad b = 6, \quad c = -7$

Now, we evaluate the expression $7a + b + c$:
$7a + b + c = 7(1) + 6 + (-7) = 6$

Thus, the correct option is (D).