$\hat{a} \cdot \hat{b}=|\hat{a}||\hat{b}| \cos \phi$
$\hat{a} \cdot \hat{b}=\cos \phi=\frac{1}{\sqrt{2}}$
$\cos \theta=\frac{(\hat{a}+\hat{b}) \cdot(\hat{a}+2 \hat{b}+2(\hat{a} \times \hat{b}))}{|\hat{a}+\hat{b}||\hat{a}+2 \hat{b}+2(\hat{a} \times \hat{b})|}$
$|\hat{a}+\hat{b}|^{2}=(\hat{a}+\hat{b}) \cdot(\hat{a}+\hat{b})$
$|\hat{a}+\hat{b}|^{2}=2+2 \hat{a} \cdot \hat{b}$
$=2+\sqrt{2}$
$\hat{a} \times \hat{b}=|\hat{a}||\hat{b}| \sin \phi \hat{n}$
$\hat{ a } \times \hat{ b }=\frac{\hat{ n }}{\sqrt{2}} \quad$ when $\hat{ n }$ is vector $\perp \hat a$ and $\hat{ b }$
let $\vec{c}=\hat{a} \times \hat{b}$
We know.
$\vec{c} \cdot \vec{a}=0$
$\overrightarrow{ c } \cdot \overrightarrow{ b }=0$
$|\hat{a}+2 \hat{b}+2 \vec{c}|^{2}$
$=1+4+\frac{(4)}{2}+4 \hat{ a } \cdot \hat{ b }+8 \hat{ b } \cdot \overrightarrow{ c }+4 \overrightarrow{ c } \cdot \hat{ a }$
$=7+\frac{4}{\sqrt{2}}=7+2 \sqrt{2}$
Now,$(\hat{a}+\hat{b}) \cdot(\hat{a}+2 \hat{b}+2 \vec{c})$
$=|\hat{a}|^{2}+2 \hat{a} \cdot \hat{b}+0+\hat{b} \cdot \hat{a}+2|\hat{b}|^{2}+0$
$=1+\frac{2}{\sqrt{2}}+\frac{1}{\sqrt{2}}+2$
$=3+\frac{3}{\sqrt{2}}$
$\cos \theta=\frac{3+\frac{3}{\sqrt{2}}}{\sqrt{2+\sqrt{2}} \sqrt{7+2 \sqrt{2}}}$
$\cos ^{2} \theta=\frac{9(\sqrt{2}+1)^{2}}{2(2+\sqrt{2})(7+2 \sqrt{2})}$
$\cos ^{2} \theta=\left(\frac{9}{2 \sqrt{2}}\right) \frac{(\sqrt{2}+1)}{(7+2 \sqrt{2})}$
$164 \cos ^{2} \theta=\frac{(82)(9)}{\sqrt{2}} \frac{(\sqrt{2}+1)}{(7+2 \sqrt{2})} \frac{(7-2 \sqrt{2})}{(7-2 \sqrt{2})}$
$=\frac{(82)}{\sqrt{2}} \frac{(9)[7 \sqrt{2}-4+7-2 \sqrt{2}]}{(41)}$
$=(9 \sqrt{2})[5 \sqrt{2}+3]$
$=90+27 \sqrt{2}$
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Column A
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Column B
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Maximum of Z
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325
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