Let → a , → b , → c be three non-coplanar vectors such that → a × →...

Taking cross product of $\overset{\to}{c}$ both side we get,

$\overset{\to}{a} \times \overset{\to}{b} = 4\overset{\to}{c} ⇒ \left(\overset{\to}{a} \times \overset{\to}{b}\right) \times \overset{\to}{c} = 0$

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

$⇒ \left(\overset{\to}{a} \cdot \overset{\to}{c}\right) . \overset{\to}{b} = \left(\overset{\to}{b} \cdot \overset{\to}{c}\right) . \overset{\to}{a}$

Since vectors are non-coplanar so,

$⇒ \left(\overset{\to}{a} \cdot \overset{\to}{c}\right) = \left(\overset{\to}{b} \cdot \overset{\to}{c}\right) = 0 . . . . . . \left(1\right)$

Similarly for $\overset{\to}{b} \times \overset{\to}{c} = 9\overset{\to}{a} ⇒ \overset{\to}{a} \cdot \overset{\to}{b} = 0 = \overset{\to}{a} \cdot \overset{\to}{c} . . . . . \left(2\right)$

So, from equation \left(1\right) \text{ and } \left(2\right)$we can say that$\overset{\to}{a}, \overset{\to}{b}, \overset{\to}{c}$are mutually$⊥$set of vectors.</p><p>Now taking modulus both side of$\overset{\to}{a} \times \overset{\to}{b} = 4\overset{\to}{c}, \overset{\to}{b} \times \overset{\to}{c} = 9\overset{\to}{a} \text{ and } \overset{\to}{c} \times \overset{\to}{a} = \alpha\overset{\to}{b}$we get,</p><p>$⇒ \left|\overset{\to}{a}\right|\left|\overset{\to}{b}\right| = 4\left|\overset{\to}{c}\right|, \left|\overset{\to}{b}\right|\left|\overset{\to}{c}\right| = 9\left|\overset{\to}{a}\right| \text{ and } \left|\overset{\to}{c}\right|\left|\overset{\to}{a}\right| = \alpha\left|\overset{\to}{b}\right|$</p><p>$⇒ \dfrac{\left|\overset{\to}{a}\right|}{\left|\overset{\to}{c}\right|} = \dfrac{4}{9}\dfrac{\left|\overset{\to}{c}\right|}{\left|\overset{\to}{a}\right|}$</p><p>$⇒ \dfrac{\left|\overset{\to}{c}\right|}{\left|\overset{\to}{a}\right|} = \dfrac{3}{2}$</p><p>$∴$If \left|a\right| = \lambda, \left|c\right| = \dfrac{3\lambda}{2} \text{ and } \left|b\right| = 6$

Now \left|\overset{\to}{a}\right| + \left|\overset{\to}{b}\right| + \left|\overset{\to}{c}\right| = 36$</p><p>$⇒ \lambda + \dfrac{3}{2}\lambda + 6 = 36$</p><p>$⇒ \lambda = 12$</p><p>Now given \left|\overset{\to}{c}\right|\left|\overset{\to}{a}\right| = \alpha\left|\overset{\to}{b}\right|$

$⇒ \alpha = \dfrac{\left|\overset{\to}{c}\right|\left|\overset{\to}{a}\right|}{\left|\overset{\to}{b}\right|} = \dfrac{3 \times 12}{2} \times \dfrac{12}{6}$

$⇒ \alpha = 36$

Note this question was bonus in jee main 27th july 2022 shift 2, so we have some modification to the originial question.