JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · Matrices

From the matrix equation, we get three equations:
$a+8b+7c=0$
$9a+2b+3c=0$
$7a+7b+7c=0$, which simplifies to $a+b+c=0$.

Solving the first two equations with the third:
From equations (1) and (3): $8b+7c=-a$ and $b+c=-a$, giving $7b+6c=0$, so $c=-\frac{7b}{6}$.

Substituting into $a+b+c=0$: $a=-b-c=-b+\frac{7b}{6}=\frac{b}{6}$, so $b=6a$ and $c=-7a$.

Using the plane equation $2a+b+c=1$: $2a+6a-7a=1$, giving $a=1$.

Therefore $b=6,c=-7$, and $7a+b+c=7+6-7=6$.