Find the unit vector which is perpendicular to the vector $5i + 2j + 6k$ and coplanar with the vectors $2i + j + k$ and $i - j + k$.

Let $\overrightarrow{c}$ be a vector perpendicular to the vectors $\overrightarrow{a}=\hat{i}+\hat{j}-\hat{k}$ and $\overrightarrow{b}=\hat{i}+2\hat{j}+\hat{k}.$ If $\overrightarrow{c}\cdot(\hat{i}+\hat{j}+3\hat{k})=8,$ then the value of $\overrightarrow{c}\cdot(\overrightarrow{a}\times\overrightarrow{b})$ is equal to ...... .

Let $2\hat{a} = \hat{b} \times \hat{c} + 2\hat{b}$. Then the sum of possible value$(s)$ of $\left| 2\hat{a} + \hat{b} + \hat{c} \right|$ is:

Find the area of the triangle with vertices $A(1,1,2), B(2,3,5)$ and $C(1,5,5)$.

Let $\lambda \in R$,$\vec{a} = \lambda \hat{i} + 2 \hat{j} - 3 \hat{k}$,and $\vec{b} = \hat{i} - \lambda \hat{j} + 2 \hat{k}$. If $((\vec{a} + \vec{b}) \times (\vec{a} \times \vec{b})) \times (\vec{a} - \vec{b}) = 8 \hat{i} - 40 \hat{j} - 24 \hat{k}$,then $|\lambda(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})|^2$ is equal to

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: