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

Let $$\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2 \hat{i}+4 \hat{j}-5 \hat{k}$$ and $$\vec{c}=x \hat{i}+2 \hat{j}+3 \hat{k}, x \in \mathbb{R}$$. If $$\vec{d}$$ is the unit vector in the direction of $$\vec{b}+\vec{c}$$ such that $$\vec{a} \cdot \vec{d}=1$$, then $$(\vec{a} \times \vec{b}) \cdot \vec{c}$$ is equal to

Let a unit vector which makes an angle of $$60^{\circ}$$ with $$2 \hat{i}+2 \hat{j}-\hat{k}$$ and an angle of $$45^{\circ}$$ with $$\hat{i}-\hat{k}$$ be $$\vec{C}$$. Then $$\vec{C}+\left(-\frac{1}{2} \hat{i}+\frac{1}{3 \sqrt{2}} \hat{j}-\frac{\sqrt{2}}{3} \hat{k}\right)$$ is: