Let $$\vec{a}=2 \hat{i}+\hat{j}+\hat{k}$$, and $$\vec{b}$$ and $$\vec{c}$$ be two nonzero vectors such that $$|\vec{a}+\vec{b}+\vec{c}|=|\vec{a}+\vec{b}-\vec{c}|$$ and $$\vec{b} \cdot \vec{c}=0$$. Consider the following two statements:
(A) $$|\vec{a}+\lambda \vec{c}| \geq|\vec{a}|$$ for all $$\lambda \in \mathbb{R}$$.
(B) $$\vec{a}$$ and $$\vec{c}$$ are always parallel.
Then,
If $((\vec{a}+\vec{b}) \times(\vec{a} \times \vec{b})) \times(\vec{a}-\vec{b})=8 \hat{i}-40 \hat{j}-24 \hat{k}$,
then $|\lambda(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})|^2$ is equal to :
If $$\overrightarrow a ,\overrightarrow b ,\overrightarrow c $$ are three non-zero vectors and $$\widehat n$$ is a unit vector perpendicular to $$\overrightarrow c $$ such that $$\overrightarrow a = \alpha \overrightarrow b - \widehat n,(\alpha \ne 0)$$ and $$\overrightarrow b \,.\overrightarrow c = 12$$, then $$\left| {\overrightarrow c \times (\overrightarrow a \times \overrightarrow b )} \right|$$ is equal to :