Let $$\vec{a} = 3\hat{i} + \hat{j} - \hat{k}$$ and $$\vec{c} = 2\hat{i} - 3\hat{j} + 3\hat{k}$$. If $$\vec{b}$$ is a vector such that $$\vec{a} = \vec{b} \times \vec{c}$$ and $$|\vec{b}|^2 = 50$$, then $$\left|72 - |\vec{b} + \vec{c}|^2\right|$$ is equal to _____.

Given $$\vec{a} = 3\hat{i} + \hat{j} - \hat{k}$$, $$\vec{c} = 2\hat{i} - 3\hat{j} + 3\hat{k}$$, $$\vec{a} = \vec{b} \times \vec{c}$$, and $$|\vec{b}|^2 = 50$$.

Key property of cross product. Since $$\vec{a} = \vec{b} \times \vec{c}$$, $$\vec{a}$$ is perpendicular to both $$\vec{b}$$ and $$\vec{c}$$.

Verify: $$\vec{a} \cdot \vec{c} = (3)(2) + (1)(-3) + (-1)(3) = 6 - 3 - 3 = 0$$ ✓

Compute magnitudes: $$|\vec{a}|^2 = 9 + 1 + 1 = 11$$

$$|\vec{c}|^2 = 4 + 9 + 9 = 22$$

Use the identity $$|\vec{b} \times \vec{c}|^2 = |\vec{b}|^2|\vec{c}|^2 - (\vec{b} \cdot \vec{c})^2$$: $$11 = 50 \times 22 - (\vec{b} \cdot \vec{c})^2$$

$$(\vec{b} \cdot \vec{c})^2 = 1100 - 11 = 1089$$

$$\vec{b} \cdot \vec{c} = \pm 33$$

Compute $$|\vec{b} + \vec{c}|^2$$: $$|\vec{b} + \vec{c}|^2 = |\vec{b}|^2 + 2(\vec{b} \cdot \vec{c}) + |\vec{c}|^2 = 50 + 2(\pm 33) + 22 = 72 \pm 66$$

So $$|\vec{b} + \vec{c}|^2 = 138$$ or $$|\vec{b} + \vec{c}|^2 = 6$$.

Find the required value: $$\left|72 - |\vec{b} + \vec{c}|^2\right| = |72 - 138| = 66 \quad \text{or} \quad |72 - 6| = 66$$

In both cases, the answer is the same.

The correct answer is 66.