Let $$\overrightarrow{\mathrm{a}}=\alpha \hat{i}+\hat{j}-\hat{k}$$ and $$\overrightarrow{\mathrm{b}}=2 \hat{i}+\hat{j}-\alpha \hat{k}, \alpha>0$$. If the projection of $$\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}$$ on the vector $$-\hat{i}+2 \hat{j}-2 \hat{k}$$ is 30, then $$\alpha$$ is equal to :

Let $$\vec{a}=\hat{i}-\hat{j}+2 \hat{k}$$ and let $$\vec{b}$$ be a vector such that $$\vec{a} \times \vec{b}=2 \hat{i}-\hat{k}$$ and $$\vec{a} \cdot \vec{b}=3$$. Then the projection of $$\vec{b}$$ on the vector $$\vec{a}-\vec{b}$$ is :

Let $$\mathrm{ABC}$$ be a triangle such that $$\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{CA}}=\overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{c}},|\overrightarrow{\mathrm{a}}|=6 \sqrt{2},|\overrightarrow{\mathrm{b}}|=2 \sqrt{3}$$ and $$\vec{b} \cdot \vec{c}=12$$. Consider the statements :

$$(\mathrm{S} 1):|(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})+(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{b}})|-|\vec{c}|=6(2 \sqrt{2}-1)$$

$$(\mathrm{S} 2): \angle \mathrm{ACB}=\cos ^{-1}\left(\sqrt{\frac{2}{3}}\right)$$

Then

Let a vector $$\overrightarrow c $$ be coplanar with the vectors $$\overrightarrow a = - \widehat i + \widehat j + \widehat k$$ and $$\overrightarrow b = 2\widehat i + \widehat j - \widehat k$$. If the vector $$\overrightarrow c $$ also satisfies the conditions $$\overrightarrow c \,.\,\left[ {\left( {\overrightarrow a + \overrightarrow b } \right) \times \left( {\overrightarrow a \times \overrightarrow b } \right)} \right] = - 42$$ and $$\left( {\overrightarrow c \times \left( {\overrightarrow a - \overrightarrow b } \right)} \right)\,.\,\widehat k = 3$$, then the value of $$|\overrightarrow c {|^2}$$ is equal to :