The value of 'a' for which the vectors
A=2i-3j+3k
B=3i+aj-7k
C=5i+3j+6k
are coplanar is

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Solution

For the vectors to be coplanar, their box product should be zero.
$⎛ ⎜ ⎝ \begin{matrix} 2 & - 3 & 3 3 & a & - 7 5 & 3 & 6 \end{matrix} ∣ ∣ ∣ ∣ = 0$
Now solving the determinant
$2 (a \times 6 - (- 7) \times 3) - (- 3) (3 \times 6 - 5 \times (- 7) + 3 (3 \times 3 - 5 \times a) = 0 ∴ a = 76$

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