Let a, b and c be three real numbers satisfying

$\left[\begin{array}{l}a\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}b\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}c\end{array}\right]\left[\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}9\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}7\\ 8\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}7\\ 7\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}7\end{array}\right]=\left[\begin{array}{l}0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\end{array}\right]$                        (1)

If the point $P(a,b,c)$ with reference to (1), lies on the plane $2x + y + z = 1$, then the value of $7a+b+c$ is

We have 

$a+8b+7c=0,9a+2b+3c=0,7a+7b+7c=0$

Subtracting the last equation from first and multiplying the

last equation by (9/7) and subtracting from the second equation
we obtain $-6a+b=0,7b+6c=0$

$\Rightarrow \frac{a}{1}=\frac{b}{6}=\frac{c}{-7}=k$ (say)

Thus, $a=k,b=6k,c=-7k$.

As $(a,b,c)=(k,6k,-7k)$ lies on $2x+y+z=1$, we get $k=1$.

Thus,  $7a+b+c=7k+6k-7k=6k=6$