Let $$\vec{a}=2 \hat{i}+5 \hat{j}-\hat{k}, \vec{b}=2 \hat{i}-2 \hat{j}+2 \hat{k}$$ and $$\vec{c}$$ be three vectors such that $$(\vec{c}+\hat{i}) \times(\vec{a}+\vec{b}+\hat{i})=\vec{a} \times(\vec{c}+\hat{i})$$. If $$\vec{a} \cdot \vec{c}=-29$$, then $$\vec{c} \cdot(-2 \hat{i}+\hat{j}+\hat{k})$$ is equal to:

Consider three vectors $$\vec{a}, \vec{b}, \vec{c}$$. Let $$|\vec{a}|=2,|\vec{b}|=3$$ and $$\vec{a}=\vec{b} \times \vec{c}$$. If $$\alpha \in\left[0, \frac{\pi}{3}\right]$$ is the angle between the vectors $$\vec{b}$$ and $$\vec{c}$$, then the minimum value of $$27|\vec{c}-\vec{a}|^2$$ is equal to:

If $$\mathrm{A}(1,-1,2), \mathrm{B}(5,7,-6), \mathrm{C}(3,4,-10)$$ and $$\mathrm{D}(-1,-4,-2)$$ are the vertices of a quadrilateral ABCD, then its area is :

For $$\lambda>0$$, let $$\theta$$ be the angle between the vectors $$\vec{a}=\hat{i}+\lambda \hat{j}-3 \hat{k}$$ and $$\vec{b}=3 \hat{i}-\hat{j}+2 \hat{k}$$. If the vectors $$\vec{a}+\vec{b}$$ and $$\vec{a}-\vec{b}$$ are mutually perpendicular, then the value of (14 cos $$\theta)^2$$ is equal to