Let $A=(\alpha, 1, 2\alpha)$,$B=(3, 1, 2)$ and $C=4\hat{i}-\hat{j}+3\hat{k}$. If $AB \times C = 6\hat{i}+9\hat{j}-5\hat{k}$,then $\alpha^2+\alpha+5=$

$A$ non-zero vector $\vec{a}$ is parallel to the line of intersection of the plane determined by the vectors $\hat{i}$ and $\hat{i}+\hat{j}$ and the plane determined by vectors $\hat{i}-\hat{j}$ and $\hat{i}+\hat{k}$. The angle between $\vec{a}$ and $(\hat{i}-2\hat{j}+2\hat{k})$ is

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

Let $\vec \alpha = 3\hat i + \hat j$ and $\vec \beta = 2\hat i - \hat j + 3\hat k.$ If $\vec \beta = \vec \beta _1 - \vec \beta _2,$ where $\vec \beta _1$ is parallel to $\vec \alpha$ and $\vec \beta _2$ is perpendicular to $\vec \alpha,$ then $\vec \beta _1 \times \vec \beta _2$ is equal to

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 $\lambda$.

If $a+2b+3c=0$ and $(a \times b)+(b \times c)+(c \times a)=\lambda(b \times c)$,then the value of $\lambda$ is equal to