Let → a = ^ i + ^ j + ^ k , → b = - ^ i - 8 ^ j + 2 ^ k and → c = 4...

Given: \overset{\to}{a} = \overset{^}{i} + \overset{^}{j} + k, \overset{\to}{b} = - \overset{^}{i} - 8\overset{^}{j} + 2\overset{^}{k} and \overset{\to}{c} = 4\overset{^}{i} + {c}_{2}\overset{^}{j} + {c}_{3}k

Also, $\overset{\to}{b} \times \overset{\to}{a} = \overset{\to}{c} \times \overset{\to}{a}$

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

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

$⇒ \overset{\to}{b} = \overset{\to}{c} + \lambda\overset{\to}{a}$

⇒ - \overset{^}{i} - 8\overset{^}{j} + 2k = \left(4\overset{^}{i} + {c}_{2}\overset{^}{j} + {c}_{3}k\right) + \lambda\left(\overset{^}{i} + \overset{^}{j} + k\right)

$⇒ \lambda + 4 = - 1, \lambda + {c}_{2} = - 8, \lambda + {c}_{3} = 2$

$⇒ \lambda = - 5, {c}_{2} = - 3, {c}_{3} = 7$

⇒ \overset{\to}{c} = 4\overset{^}{i} - 3\overset{^}{j} + 7k

Now, finding angle we get,

cos\theta = \dfrac{\left(3\overset{^}{i} + 4\overset{^}{j} + \overset{^}{k}\right) \cdot \left(4\overset{^}{i} - 3\overset{^}{j} + 7\overset{^}{k}\right)}{\left|\left(3\overset{^}{i} + 4\overset{^}{j} + \overset{^}{k}\right)\right| ⋅ \left|\left(4\overset{^}{i} - 3\overset{^}{j} + 7\overset{^}{k}\right)\right|}

$⇒ cos\theta = \dfrac{12 - 12 + 7}{\sqrt{26} ⋅ \sqrt{74}}$

$⇒ cos\theta = \dfrac{7}{\sqrt{26} ⋅ \sqrt{74}}$

$⇒ cos\theta = \dfrac{7}{2\sqrt{481}}$

$⇒ sec\theta = \dfrac{2\sqrt{481}}{7}$

$⇒ \sec^{2}\theta = \dfrac{4 \times 481}{49}$

$⇒ 1 + \tan^{2}\theta = \dfrac{1924}{49}$

$⇒ \tan^{2}\theta = \dfrac{1875}{49}$

$⇒ \tan^{2}\theta \approx 38 . 26$

$⇒ \left[\tan^{2}\theta\right] = \left[38 . 26\right] = 38$