(1)
Again,
(
)
b
a
b
c
⋅
+
+
r
r
r
r  = 0
or
a b
b c
⋅
+
⋅
r
r r
r
 =
2
16
−b
= −
r (2)
Similarly
a c
b c
⋅
+
r⋅
r r
r  = – 4 (3)
Adding (1), (2) and (3), we have
2 (
)
a b
b
a
c
c
⋅
+
⋅ +
⋅
r
r
r
r
r
r
 = – 21
or
2μ = – 21, i e