Let → a = α ^ i + 2 ^ j - ^ k and → b = - 2 ^ i + α ^ j + ^ k ,...

Given,

\overset{\to}{a} = \alpha\overset{^}{i} + 2\overset{^}{j} - \overset{^}{k}, \overset{\to}{b} = - 2\overset{^}{i} + \alpha\overset{^}{j} + \overset{^}{k},

Area of parallelogram = \left|\overset{^}{a} \times \overset{^}{b}\right|</p><p>\overset{^}{a} \times \overset{^}{b} = \sqrt{{\left(\alpha + 2\right)}^{2} + {\left(\alpha - 2\right)}^{2} + {\left({\alpha}^{2} + 4\right)}^{2}}

Given \left|\overset{^}{a} \times \overset{^}{b}\right|$= \sqrt{15 \left({\alpha}^{2} + 4\right)}$$</p><p>So,$2\left({\alpha}^{2} + 4\right)$+ {\left({\alpha}^{2} + 4\right)}$^{2} = 15 \left({\alpha}^{2} + 4\right)$</p><p>$⇒ {\left({\alpha}^{2} + 4\right)}^{2} = 13 \left({\alpha}^{2} + 4\right)

$⇒ {\alpha}^{2} + 4 = 13 ∴ {\alpha}^{2} = 9$

Now, $2{\left|\overset{\to}{a}\right|}^{2} + \left(\overset{\to}{a} \cdot \overset{\to}{b}\right){\left|\overset{\to}{b}\right|}^{2}$

${\left|\overset{\to}{a}\right|}^{2} = {\alpha}^{2} + 4 + 1 = {\alpha}^{2} + 5$

${| \overset{\to}{b} |}^{2} = 4 + {\alpha}^{2} + 1 = {\alpha}^{2} + 5$

$\overset{\to}{a} \cdot \overset{\to}{b} = - 2\alpha + 2\alpha - 1 = - 1$

$∴ 2{\left|\overset{\to}{a}\right|}^{2} + \left(\overset{\to}{a} \cdot \overset{\to}{b}\right){\left|\overset{\to}{b}\right|}^{2}$

$2\left({\alpha}^{2} + 5\right) - 1\left({\alpha}^{2} + 5\right) = {\alpha}^{2} + 5 = 14$