If the points having the position vectors $-\hat{i}+4 \hat{j}-4 \hat{k}$,$3 \hat{i}+2 \hat{j}-5 \hat{k}$,$-3 \hat{i}+8 \hat{j}-5 \hat{k}$ and $-3 \hat{i}+2 \hat{j}+\lambda \hat{k}$ are coplanar,then $\lambda=$

If the vectors $2 \hat{i}+3 \hat{j}+4 \hat{k}$,$2 \hat{i}+\hat{j}-\hat{k}$ and $\lambda \hat{i}-\hat{j}+2 \hat{k}$ are coplanar,then the value of $\lambda$ is:

Let $S$ be the set of all $(\lambda, \mu)$ for which the vectors $\lambda \hat{i} - \hat{j} + \hat{k}$,$\hat{i} + 2\hat{j} + \mu \hat{k}$ and $3\hat{i} - 4\hat{j} + 5\hat{k}$,where $\lambda - \mu = 5$,are coplanar,then $\sum_{(\lambda, \mu) \in S} 80(\lambda^2 + \mu^2)$ is equal to :

Let $\overrightarrow{a}$ be a unit vector, $\overrightarrow{b} = 2\hat{i} + \hat{j} - \hat{k}$ and $\overrightarrow{c} = \hat{i} + 3\hat{k}$. Then, the maximum value of $[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]$ is

The lines $\overline{r}=\overline{a}+\lambda(\overline{b} \times \overline{c})$ and $\overline{r}=\overline{c}+\mu(\overline{a} \times \overline{b})$ will intersect if

Let the position vectors of the points $A, B, C$ and $D$ be $5\hat{i}+5\hat{j}+2\lambda\hat{k}$,$\hat{i}+2\hat{j}+3\hat{k}$,$-2\hat{i}+\lambda\hat{j}+4\hat{k}$ and $-\hat{i}+5\hat{j}+6\hat{k}$. Let the set $S = \{\lambda \in \mathbb{R} : \text{The points } A, B, C \text{ and } D \text{ are coplanar}\}$. Then $\sum_{\lambda \in S}(\lambda+2)^2$ is equal to