Let → a = - ^ i - ^ j + ^ k , → a · → b = 1 and → a × → b = ^ i - ^...

Let

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

Then,

\overset{\to}{a} \times \overset{\to}{b} = \left|\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ - 1 & - 1 & 1 \\ x & y & z\end{cases}\right|

⇒ \overset{\to}{a} \times \overset{\to}{b} = - \left(y + z\right)\overset{^}{i} + \left(x + z\right)\overset{^}{j} + \left(x - y\right)\overset{^}{k}

So,

- \left(y + z\right)\overset{^}{i} + \left(x + z\right)\overset{^}{j} + \left(x - y\right)\overset{^}{k} = \overset{^}{i} - \overset{^}{j} + 0\overset{^}{k}

Hence,

$y + z = - 1x + z = - 1x - y = 0 ⇒ x = y$

Also,

$\overset{\to}{a} \cdot \overset{\to}{b} = 1$

$⇒ - x - y + z = 1$

$⇒ - 2x + z = 1$

$⇒ z = 2x + 1$

$⇒ - 1 - x = 2x + 1$

$⇒ 3x = - 2 ⇒ x = - \dfrac{2}{3}$

So, $y = - \dfrac{2}{3}; z = - \dfrac{1}{3}$

Hence,

\overset{\to}{b} = - \dfrac{2}{3}\overset{^}{i} - \dfrac{2}{3}\overset{^}{j} - \dfrac{1}{3}\overset{^}{k}

⇒ 6\overset{\to}{b} = - 4\overset{^}{i} - 4\overset{^}{j} - 2\overset{^}{k}

So,

\overset{\to}{a} - 6\overset{\to}{b} = 3\left(\overset{^}{i} + \overset{^}{j} + \overset{^}{k}\right)