Let → a be a non-zero vector parallel to the line of intersection...

Given,

$\overset{\to}{a}$ be a non-zero vector parallel to the line of intersection of the two planes described by \overset{^}{i} + \overset{^}{j}, \overset{^}{i} + \overset{^}{k},

So finding the normal vector to the above plane, we get,

\overset{\to}{n}1 = \left|\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ 1 & 1 & 0 \\ 1 & 0 & 1\end{cases}\right| = \overset{^}{i} - \overset{^}{j} - \overset{^}{k}

And normal vector to planes \overset{^}{i} - \overset{^}{j}, \overset{^}{j} - \overset{^}{k} will be,

\overset{\to}{n}2 = \left|\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ 1 & - 1 & 0 \\ 0 & 1 & - 1\end{cases}\right| = \overset{^}{i} + \overset{^}{j} + \overset{^}{k}

Now $\overset{\to}{a} = \lambda\left(\overset{\to}{n}1 \times \overset{\to}{n}2\right)$

So, direction ratio of \overset{\to}{a} = \left|\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ 1 & - 1 & - 1 \\ 1 & 1 & 1\end{cases}\right|= - 2\overset{^}{j} + 2\overset{^}{k}

Hence, direction ratio of $\overset{\to}{a} = \left(0, - 2, 2\right) = \left(0, - 1, 1\right)$

Now given, \overset{\to}{b} = 2\overset{^}{i} - 2\overset{^}{j} + \overset{^}{k}

And \overset{\to}{a} = \lambda ( - \overset{^}{j} + \overset{^}{k} )

So, $\overset{\to}{a} \cdot \overset{\to}{b} = 6 = \lambda ( 2 + 1 )$

$⇒ \lambda = 2$

∴ \overset{\to}{a} = - 2\overset{^}{j} + 2\overset{^}{k}

Now we know that, $\overset{\to}{a} \cdot \overset{\to}{b} = | \overset{¯}{a} | | b | cos\theta$

$⇒ 6 = 2\sqrt{2} \times 3cos\theta$

$⇒ cos\theta = \dfrac{1}{\sqrt{2}}$

$⇒ \theta = \dfrac{\pi}{4}$

Now finding, \left|\overset{\to}{a} \times \overset{\to}{b}\right| = \left|\overset{\to}{a}\right|\left|\overset{\to}{b}\right|sin\theta = 2\sqrt{2} \times 3 \times \dfrac{1}{\sqrt{2}} = 6$</p><p>Hence, ordered pair \left(\theta, \left|\overset{\to}{a} \times \overset{\to}{b}\right|\right) = \left(\dfrac{\pi}{4}, 6\right)$