A- b- c- are non coplanar vectors and p- q- r- are defined as p-b-c- b- c- a- - q-c-a- c- a- b- - r-a-b- a- b- c- - then a-b- -p- b-c- -q- c-a- -r- is equal to

The correct option is A 3 Given #160; p⃗=b⃗×c⃗[b⃗c⃗a⃗] q⃗=c⃗×a⃗[c⃗a⃗b⃗] r⃗=a⃗×b⃗[a⃗b⃗c⃗] (a⃗+b⃗)·p⃗+(b⃗+c⃗)·q⃗+(c⃗+a⃗)·r⃗ (a⃗+b⃗ ...

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Definition of a Determinant

a⃗, b⃗, c⃗ ar...

Question

$\to a, \to b, \to c$ are non coplanar vectors and $\to p, \to q, \to r$ are defined as $\to p = \frac{\to b \times \to c}{[\begin{matrix} \to b & \to c & \to a \end{matrix}]}$ , $\to q = \frac{\to c \times \to a}{[\begin{matrix} \to c & \to a & \to b \end{matrix}]}$ , $\to r = \frac{\to a \times \to b}{[\begin{matrix} \to a & \to b & \to c \end{matrix}]}$ , then $(\to a + \to b) . \to p + (\to b + \to c) . \to q + (\to c + \to a) . \to r$ is equal to

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\to a, \to b, \to c

\to p, \to q, \to r

(\to a \times \to p) + (\to b \times \to q) + (\to c \times \to r)

\to a, \to b, \to c

\to p = \frac{\to b \times \to c}{[\to a \to b \to c]}

\to q = \frac{\to c \times \to a}{[\to a \to b \to c]}

\to r = \frac{\to a \times \to b}{[\to a \to b \to c]}

(2 \to a + 3 \to b + 4 \to c) . \to p + (2 \to b + 3 \to c + 4 \to a) . \to q +

(2 \to c + 3 \to a + 4 \to b) . \to r =

¯ ¯ ¯ a, ¯ ¯ b, ¯ ¯ c

¯ ¯ ¯ p, ¯ ¯ ¯ q, ¯ ¯ ¯ r

\to p = \frac{\to b \times \to c}{[\begin{matrix} \to a & \to b & \to c \end{matrix}]}, \to q = \frac{\to c \times \to a}{[\begin{matrix} \to a & \to b & \to c \end{matrix}]}, \to r = \frac{\to a \times \to b}{[\begin{matrix} \to a & \to b & \to c \end{matrix}]}

(\to a + \to b + \to c) . (\to p + \to q + \to r)

\to a, \to b, \to c

\to r

(\to a \times \to b) \times (\to r \times \to c) + (\to b \times \to c) \times (\to r \times \to a) + (\to c \times \to a) \times (\to r \times \to b) =

\to a, \to b

\to c

\to p, \to q

\to r

\to p = \to a + \to b - 2 \to c, \to q = 3 \to a - 2 \to b + \to c

\to r = \to a - 4 \to b + 2 \to c

\to a, \to b

\to c

V_{1}

\to p, \to q

\to r

V_{2}

V_{2} : V_{1} =

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