If $\vec{a}=\vec{b}+\vec{c}, \vec{b} \times \vec{d}=\vec{0}, \vec{c} . \vec{d}=0$, then the vector $\frac{\vec{d} \times(\vec{a} \times \vec{d})}{|\vec{d}|^2}$ is always equal to:

$\vec{c}$

$\vec{a}=\vec{b}+\vec{c}, \vec{b} \times \vec{d}=0, \vec{c} . \vec{d}=0$

$\Rightarrow \vec{a} \times \vec{d}=\vec{b} \times \vec{d}+\vec{c} \times \vec{d}=\vec{c} \times \vec{d}$

$\Rightarrow \vec{d} \times(\vec{a} \times \vec{d})=\vec{d} \times(\vec{c} \times \vec{d})$

$=(\vec{d} . \vec{d}) \vec{c}-(\vec{c} . \vec{d}) \vec{d}$

$=|\vec{d}|^2 \vec{c}$

$\Rightarrow \frac{\vec{d} \times(\vec{a} \times \vec{d})}{|\vec{d}|^2}=\vec{c}$

Hence (4) is correct answer.