If $\to a, \to b a n d \to c$ are non-coplanar vectors, then the following vectors are coplaner-

\to a + 2 \to b + 3 \to c, - 2 \to a + 3 \to b - 4 \to c, \to a - 3 \to b + 5 \to c

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3 \to a - 7 \to b - 4 \to c, 3 \to a - 2 \to b + \to c, \to a + \to b + 2 \to c

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\to a - 2 \to b + 3 \to c, - 2 \to a + 3 \to b - 4 \to c, \to a - \to b + 2 \to c

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7 \to a - 8 \to b + 9 \to c, 3 \to a + 20 \to b + 5 \to c, 5 \to a + 6 \to b + 7 \to c

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Solution

The correct option is C $3 \to a - 7 \to b - 4 \to c, 3 \to a - 2 \to b + \to c, \to a + \to b + 2 \to c$
For any three vectors to be co planar, their scalar triple product must be equal to zero.
Triple product can be found by writing down the numbers as a matrix and calculating its determinant.
Lets check option B,
$3 \to a - 7 \to b - 4 \to c 3 \to a - 2 \to b + \to c \to a + \to b + 2 \to c$
Now, lets find scalar triple product,
$∣ ∣ ∣ ∣ \begin{matrix} 3 & - 7 & - 4 3 & - 2 & 1 1 & 1 & 2 \end{matrix} ∣ ∣ ∣ ∣ = 3 [(- 2) (2) - (1) (1)] - (- 7) [(3) (2) - (1) (1)] - 4 [3 (1) - 1 (- 2)] = 3 [- 4 - 1] + 7 [6 - 1] - 4 [3 + 2] = 3 (- 5) + 7 (5) - 4 (5) = - 15 + 35 - 20 = 0$
As the scalar triple product is zero, the three vectors
in option B are coplanar.

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