\[ (\sqrt{2})^x = 2^x \implies x = 0 \implies \alpha = 1 \]
\[ z = \frac{\pi}{4}(1 + i)^4 \]
\[ = \frac{-\pi i}{2} \left[ \sqrt{\frac{\pi - \pi i - \sqrt{\pi}}{\pi + 1}} + \sqrt{\frac{\pi - i - \pi i - \sqrt{\pi}}{1 + \pi}} \right] \]
\[ = \frac{-\pi i}{2} \left( 1 + 4i + 6i^2 + 4i^3 + 1 \right) \]
\[ = 2\pi i \beta = \frac{2\pi}{\pi/2} = 4 \]
Distance from \( (1, 4) \) to \( 4x - 3y = 7 \) will be:
\[ \frac{15}{5} = 3 \]
Let \( \vec{a} = 3\hat{i} + \hat{j} - 2\hat{k} \), \( \vec{b} = 4\hat{i} + \hat{j} + 7\hat{k} \), and \( \vec{c} = \hat{i} - 3\hat{j} + 4\hat{k} \) be three vectors.
If a vector \( \vec{p} \) satisfies \( \vec{p} \times \vec{b} = \vec{c} \times \vec{b} \) and \( \vec{p} \cdot \vec{a} = 0 \), then \( \vec{p} \cdot (\hat{i} - \hat{j} - \hat{k}) \) is equal to