Let a → = i ^ + 4 j ^ + 2 k ^ , b → = 3 i ^ - 2 j ^ + 7 k ^ and c → = 2 i ^ - j ^ + 4 k ^ …

We have been given that

$\overrightarrow{d}\times \overrightarrow{b}=\overrightarrow{c}\times \overrightarrow{b}$

$\Rightarrow (\overrightarrow{d}-\overrightarrow{c})\times \overrightarrow{b}=0$

$\Rightarrow \overrightarrow{d}-\overrightarrow{c}=\lambda \overrightarrow{b} (\because \overrightarrow{d}\ne \overrightarrow{c})$

$\Rightarrow \overrightarrow{d}=\overrightarrow{c}+\lambda \overrightarrow{b}$

It is given that $\overrightarrow{a}·\overrightarrow{d}=24$ 

$\Rightarrow \overrightarrow{a}·\overrightarrow{c}+\lambda \overrightarrow{b}·\overrightarrow{a}=24$

$\Rightarrow 6+\lambda (3-8+14)=24$

$\Rightarrow 9\lambda =18$

$\Rightarrow \lambda =2$

$\therefore \overrightarrow{d}=\overrightarrow{c}+2\overrightarrow{b}$

$\Rightarrow \overrightarrow{d}=8\hat{\mathrm{i}}-5\hat{\mathrm{j}}+18\hat{\mathrm{k}}$

$|\overrightarrow{d}{|}^2=64+25+324$$=413$

Hence, this is the correct option.