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 $\alpha^2+\alpha+5=$

If $a=\hat{i}+\hat{j}$ and $b=3 \hat{i}-2 \hat{j}$, then the vector $r$ satisfying the equations $r \times a=b \times a$ and $r \times b=a \times b$ is

$A$ unit vector perpendicular to the plane containing the vectors $\hat{i}+2\hat{j}+\hat{k}$ and $-2\hat{i}+\hat{j}+3\hat{k}$ is

If $\overrightarrow{a}=2 \hat{i}-5 \hat{j}+8 \hat{k}$ and $\overrightarrow{b}=7 \hat{i}-5 \hat{j}+3 \hat{k}$ are two vectors and $(2 \overrightarrow{a}-3 \overrightarrow{b}) \times(4 \overrightarrow{a}+\overrightarrow{b})=x \hat{i}+y \hat{j}+z \hat{k}$,then $x+y+z=$

The unit vectors perpendicular to the plane determined by the points $A(1, -1, 2)$,$B(2, 0, -1)$,and $C(0, 2, 1)$ are:

$A, B, C, D$ are any $4$ points and $|\overline{AB} \times \overline{CD} + \overline{BC} \times \overline{AD} + \overline{CA} \times \overline{BD}| = \lambda$ (Area of $\triangle ABC$). Then $\lambda = $