Let → a = ^ i + 2 ^ j + 3 ^ k , → b = ^ i - ^ j + 2 ^ k and → c = 5...

Given,

$\overset{\to}{r} \times \overset{\to}{b} = \overset{\to}{c} \times \overset{\to}{b}$

$⇒ \overset{\to}{r} \times \overset{\to}{b} - \overset{\to}{c} \times \overset{\to}{b} = 0$

\left(\overset{\to}{r} - \overset{\to}{c}\right) \times \overset{\to}{b} = 0$</p><p>Hence, we can say that$\overset{\to}{r} - \overset{\to}{c}$is parallel to vector$\overset{\to}{b}$,</p><p>Now let ,$\overset{\to}{r} - \overset{\to}{c} = \lambda\overset{\to}{b}$</p><p>$⇒ \overset{\to}{r} = \overset{\to}{c} + \lambda\overset{\to}{b} . . . . . . \left(1\right)$</p><p>$∵ \overset{\to}{r} \cdot \overset{\to}{a} = 0 ⇒ \overset{\to}{c} \cdot \overset{\to}{a} + \lambda\left(\overset{\to}{b} \cdot \overset{\to}{a}\right) = 0$</p><p>$⇒ \lambda = \dfrac{- \overset{\to}{c} \cdot \overset{\to}{a}}{\overset{\to}{b} \cdot \overset{\to}{a}} = \dfrac{- 8}{5}$</p><p>Now putting the value of$\lambda$in equation \left(1\right)$ we get,

\overset{\to}{r} = \overset{\to}{c} - \dfrac{8}{5}\overset{\to}{b} = \left(5\overset{^}{i} - 3\overset{^}{j} + 3\overset{^}{k}\right) - \dfrac{8}{5}\left(\overset{^}{i} - \overset{^}{j} + 2\overset{^}{k}\right)

⇒ 5\overset{\to}{r} = 17\overset{^}{i} - 7\overset{^}{j} - \overset{^}{k}

$⇒ 25{\left|\overset{\to}{r}\right|}^{2} = {\left(17\right)}^{2} + {\left(- 7\right)}^{2} + {\left(- 1\right)}^{2}$

$⇒ 25{\left|\overset{\to}{r}\right|}^{2} = 289 + 49 + 1 = 339$