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

Consider the vectors $u = a \hat{i} + b \hat{j} + c \hat{k}$,$v = a^2 \hat{i} + b^2 \hat{j} + c^2 \hat{k}$ and $w = a^3 \hat{i} + b^3 \hat{j} + c^3 \hat{k}$. These vectors are coplanar if and only if

For non-zero vectors $\vec{a}, \vec{b}, \vec{c}$,the condition $|(\vec{a} \times \vec{b}) \cdot \vec{c}| = |\vec{a}||\vec{b}||\vec{c}|$ holds if and only if:

If the points whose position vectors are $3i - 2j - k,$ $2i + 3j - 4k,$ $-i + j + 2k,$ and $4i + 5j + \lambda k$ lie on a plane,then $\lambda = $

The value of $[\vec{a}-\vec{b} \quad \vec{b}-\vec{c} \quad \vec{c}-\vec{a}]$ is equal to

If $\overline{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}, \overline{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}$ and $\overline{c}=c_1 \hat{i}+c_2 \hat{j}+c_3 \hat{k}$ are non-zero non-coplanar vectors and $m$ is a non-zero scalar such that $[m\overline{a}+\overline{b} \quad m\overline{b}+\overline{c} \quad m\overline{c}+\overline{a}] = 28[\overline{a} \quad \overline{b} \quad \overline{c}]$,then the value of $m$ is: