If $$\vec{a}$$, $$\vec{b}$$, $$\vec{c}$$ are three non-zero vectors and $$\hat{n}$$ is a unit vector perpendicular to $$\vec{c}$$ such that $$\vec{a} = \alpha\vec{b} - \hat{n}$$, $$\alpha \neq 0$$ and $$\vec{b} \cdot \vec{c} = 12$$, then $$\vec{c} \times \vec{a} \times \vec{b}$$ is equal to:

To find the value of the expression, we use the vector triple product identity:

$$\vec{c} \times (\vec{a} \times \vec{b}) = (\vec{c} \cdot \vec{b})\vec{a} - (\vec{c} \cdot \vec{a})\vec{b}$$

We are given the following conditions:

$$\vec{b} \cdot \vec{c} = 12$$

$$\vec{a} = \alpha\vec{b} - \hat{n}$$

$$\hat{n} \cdot \vec{c} = 0 \quad \text{(since } \hat{n} \text{ is perpendicular to } \vec{c}\text{)}$$

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Step 1: Find the value of $$\vec{c} \cdot \vec{a}$$

Taking the dot product with $$\vec{c}$$ on both sides of the vector equation $$\vec{a} = \alpha\vec{b} - \hat{n}$$:

$$\vec{c} \cdot \vec{a} = \vec{c} \cdot (\alpha\vec{b} - \hat{n})$$

$$\vec{c} \cdot \vec{a} = \alpha(\vec{c} \cdot \vec{b}) - \vec{c} \cdot \hat{n}$$

Substituting the given values $$\vec{b} \cdot \vec{c} = 12$$ and $$\vec{c} \cdot \hat{n} = 0$$:

$$\vec{c} \cdot \vec{a} = \alpha(12) - 0 = 12\alpha$$

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Step 2: Substitute these values into the vector triple product expression

$$\vec{c} \times (\vec{a} \times \vec{b}) = (12)\vec{a} - (12\alpha)\vec{b}$$

$$\vec{c} \times (\vec{a} \times \vec{b}) = 12(\vec{a} - \alpha\vec{b})$$

From the given relation, we know that $$\vec{a} = \alpha\vec{b} - \hat{n} \implies \vec{a} - \alpha\vec{b} = -\hat{n}$$. Substituting this into the simplified expression:

$$\vec{c} \times (\vec{a} \times \vec{b}) = 12(-\hat{n}) = -12\hat{n}$$

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Step 3: Calculate the magnitude of the resulting vector

Taking the absolute magnitude on both sides:

$$\left|\vec{c} \times (\vec{a} \times \vec{b})\right| = |-12\hat{n}| = 12|\hat{n}|$$

Since $$\hat{n}$$ is a unit vector, its magnitude $$|\hat{n}| = 1$$:

$$\left|\vec{c} \times (\vec{a} \times \vec{b})\right| = 12 \times 1 = 12$$

Therefore, the magnitude of the expression is equal to 12.