$\overrightarrow{b} = \hat{i} - \hat{j} +\hat{k} \Rightarrow \overrightarrow{|b|} = \sqrt{1^{2} + (1)^{2} + (-1)^{2}} = \sqrt{3}$
$\overrightarrow{a}.\overrightarrow{b} = \overrightarrow{|a|} \overrightarrow{|b|} \cos\Theta$
$\Rightarrow 1-1-1 = \sqrt{3}.\sqrt{3} \cos \theta \Rightarrow -1 = 3 \cos\theta$
$\Rightarrow \cos\theta = -\frac{1}{3} \Rightarrow \theta = \cos^{-1} \bigg(-\frac{1}{3}\bigg)$
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