Answer:
Option B
Explanation:
$r.a= \alpha( a.b \times c)+\beta(a.c \times a)+\gamma(a.a \times b)$
$=\alpha[abc]+0+0$ Similarly
$r.b=\beta [a bc]$ and $r.c=\gamma [abc]$
$\therefore$ $\frac{1}{2} r.(a+b+c)=\frac{1}{2} (r.a+r.b+r.c)$
$\frac{1}{2} (\alpha+\beta+\gamma)[abc]$
=$\frac{1}{2}(\alpha+\beta+\gamma)\times 2=\alpha+\beta+\gamma$