Let → a = ^ i - 2 ^ j + 3 ^ k , → b = ^ i + ^ j + ^ k and → c be a...

Given, \overset{\to}{a} = \overset{^}{i} - 2\overset{^}{j} + 3\overset{^}{k}, \overset{\to}{b} = \overset{^}{i} + \overset{^}{j} + \overset{^}{k}

And $\overset{\to}{a} \times ( \overset{\to}{b} + \overset{\to}{c} ) = \overset{\to}{0}$,

Let \overset{\to}{c} = x\overset{^}{i} + y\overset{^}{j} + z\overset{^}{k}

So, $\overset{\to}{a} \times \left(\overset{\to}{b} + \overset{\to}{c}\right) = \overset{\to}{0} ⇒ \overset{\to}{a} \times \overset{\to}{b} + \overset{\to}{a} \times \overset{\to}{c} = \overset{\to}{0}$

$⇒ \overset{\to}{a} \times \overset{\to}{b} = - \overset{\to}{a} \times \overset{\to}{c} ⇒ \overset{\to}{a} \times \overset{\to}{b} = \overset{\to}{c} \times \overset{\to}{a}$

⇒ \left|\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ 1 & - 2 & 3 \\ 1 & 1 & 1\end{cases}\right| = \left|\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ x & y & z \\ 1 & - 2 & 3\end{cases}\right|

⇒ - 5\overset{^}{i} + 2\overset{^}{j} + 3\overset{^}{k} = \overset{^}{i}\left(3y + 2z\right) + \overset{^}{j}\left(z - 3x\right) + \overset{^}{k}\left(- 2x - y\right)

On comparing we get,

$3y + 2z = - 5$ .....(i)

$z - 3x = 2$ ............(ii)

$- 2x - y = 3$ ........(iii)

On solving equation (i), (ii) and (iii) we get,

$x = \dfrac{- 1}{6}, y = \dfrac{- 8}{3}, z = \dfrac{3}{2}$

So, \overset{\to}{c} = \dfrac{- 1}{6}\overset{^}{i} + \dfrac{- 8}{3}\overset{^}{j} + \dfrac{3}{2}\overset{^}{k}

So, \overset{\to}{c} . \overset{\to}{a} = \left(\dfrac{- 1}{6}\overset{^}{i} + \dfrac{- 8}{3}\overset{^}{j} + \dfrac{3}{2}\overset{^}{k}\right) . \left(\overset{^}{i} - 2\overset{^}{j} + 3\overset{^}{k}\right) = \dfrac{29}{3}

So, $3\overset{\to}{c} . \overset{\to}{a} = 29$

Note: This question appeared in JEE Main 2022, 29th june shift 2. The question was incorrect, so it is modified to make it correct.