The position vectors of the points $A, B$ and $C$ are $i + j, j + k$ and $k + i$ respectively. The vector area of the $\Delta ABC = \pm \frac{1}{2} \vec{\alpha}$ where $\vec{\alpha} = $

$\vec{a}=\hat{i}+\hat{j}-2 \hat{k}$,$\vec{b}=\hat{i}-2 \hat{j}+\hat{k}$ and $\vec{c}=2 \hat{i}+\hat{j}-\hat{k}$ are three vectors. If $\vec{d}$ is a normal to the plane of $\vec{a}$ and $\vec{b}$ and $\vec{d} \cdot \vec{c}=2$,then $|\vec{d}|=$

$a, b, c, d$ are coplanar vectors,then $(a \times b) \times (c \times d)$ is equal to

If $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2 \hat{i}-\hat{j}+3 \hat{k}$ and $\vec{c}=\hat{i}-\hat{j}$ and if $6 \hat{i}+2 \hat{j}+3 \hat{k}=\lambda_1(\vec{a} \times \vec{b})+\lambda_2(\vec{b} \times \vec{c})+\lambda_3(\vec{c} \times \vec{a})$,then $(\lambda_1, \lambda_2, \lambda_3)=$

$(2a + 3b) \times (5a + 7b) = $

If $A, B, C, D$ are four points in space,then $|\overline{AB} \times \overline{CD} + \overline{BC} \times \overline{AD} + \overline{CA} \times \overline{BD}| = \lambda \times (\text{Area of } \Delta ABC)$. Find $\lambda$.