Let $ABC$ be a triangle of area $15 \sqrt{2}$ and the vectors $\overrightarrow{AB}=\hat{i}+2 \hat{j}-7 \hat{k}$,$\overrightarrow{BC}=a \hat{i}+b \hat{j}+c \hat{k}$ and $\overrightarrow{AC}=6 \hat{i}+d \hat{j}-2 \hat{k}$,where $d>0$. Then the square of the length of the largest side of the triangle $ABC$ is:

For some real number $\lambda$,if the area of the triangle having $\vec{a}=3 \hat{i}-\hat{j}+\lambda \hat{k}$ and $\vec{b}=\lambda \hat{i}+\hat{j}-3 \hat{k}$ as two of its sides is $\frac{\sqrt{195}}{2}$,then the number of distinct possible values of $\lambda$ is

If $a = 2i + k$,$b = i + j + k$ and $c = 4i - 3j + 7k$. If $d \times b = c \times b$ and $d \cdot a = 0$,then $d$ is equal to:

Let $\overrightarrow{a} = \hat{i} + 2\hat{j} + \hat{k}$,$\overrightarrow{b} = 3\hat{i} - 3\hat{j} + 3\hat{k}$,$\overrightarrow{c} = 2\hat{i} - \hat{j} + 2\hat{k}$ and $\overrightarrow{d}$ be a vector such that $\overrightarrow{b} \times \overrightarrow{d} = \overrightarrow{c} \times \overrightarrow{d}$ and $\overrightarrow{a} \cdot \overrightarrow{d} = 4$. Then $|(\overrightarrow{a} \times \overrightarrow{d})|^2$ is equal to . . . . . . .

Let $\bar{a}=2 \hat{i}+\hat{j}-2 \hat{k}$,$\bar{b}=\hat{i}+\hat{j}$ and $\bar{c}$ be a vector such that $|\bar{c}-\bar{a}|=4$,$|(\bar{a} \times \bar{b}) \times \bar{c}|=3$ and the angle between $\bar{c}$ and $\bar{a} \times \bar{b}$ is $\frac{\pi}{6}$,then $\bar{a} \cdot \bar{c}$ is equal to

If $a \times b = b \times c \ne 0,$ where $a, b$ and $c$ are coplanar vectors,then for some scalar $k$