Correct option is (A) 0
\(\vec a = \hat i + \hat j - \hat k\)
\(\vec c = 2\hat i - 3\hat j + 2\hat k\)
Let \(\vec b = b_1\hat i + b_2\hat j + b_3 \hat k\)
Then
\(\vec b \times \vec c = \begin{vmatrix} \hat i&\hat j & \hat k\\b_1 &b_2 &b_3\\2 &-3&2\end{vmatrix}\)
\(= \hat i (2 b_2 + 3b_3) - \hat j(2b_1 - 2b_3) + \hat k (-3b_1 - 2b_2)\)
\(= \hat a\) (Given)
\(\therefore 2b_2 + 3b_3 = 1\) .....(1)
\(2b_3 - 2b_1 = 1\)
⇒ \(b_3 = \frac12 + b_1\) .....(2)
\(3b_1 - 2b_2 = -1 \)
⇒ \(3b_1 + 2b_2 = 1\) ....(3)
\(\therefore \) From (1), we get
\(2b_2 + \frac32 + 3{b_1} = 1\)
⇒ \(2b_2 + 3b_1 = 1 - \frac32 = \frac{-1}2\)
But
\(2b_2 + 3b_1 = 1\) (From(3))
\(\therefore\) Such \(\vec b\) is not possible.