\(1\,\mu\)C charge moving with velocity \(\vec{v} = \left(\hat{i} - 2\hat{j} + 3\hat{k}\right)\) m/s in the region of magnetic field \(\vec{B} = \left(2\hat{i} + 3\hat{j} - 5\hat{k}\right)\) T. The magnitude of force acting on it is \(\sqrt{\alpha} \times 10^{-6}\) N. The value of \(\alpha\) is _______.

The magnetic force acting on a moving charge is given by \(\vec{F} = q(\vec{v} \times \vec{B})\).

Given:
\(q = 1 \, \mu\text{C} = 10^{-6} \text{ C}\)
\(\vec{v} = \hat{i} - 2\hat{j} + 3\hat{k}\)
\(\vec{B} = 2\hat{i} + 3\hat{j} - 5\hat{k}\)

Calculating the cross product \(\vec{v} \times \vec{B}\):
\(\vec{v} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & 3 & -5 \end{vmatrix}\)

\(\vec{v} \times \vec{B} = \hat{i}(10 - 9) - \hat{j}(-5 - 6) + \hat{k}(3 + 4)\)
\(\vec{v} \times \vec{B} = \hat{i} + 11\hat{j} + 7\hat{k}\)

The magnitude of \(\vec{v} \times \vec{B}\) is:
\(|\vec{v} \times \vec{B}| = \sqrt{1^2 + 11^2 + 7^2} = \sqrt{1 + 121 + 49} = \sqrt{171}\)

The magnitude of the force is:
\(|\vec{F}| = q |\vec{v} \times \vec{B}| = 10^{-6} \times \sqrt{171} = \sqrt{171} \times 10^{-6} \text{ N}\)

Comparing this with the given expression \(\sqrt{\alpha} \times 10^{-6} \text{ N}\), we get:
\(\alpha = 171\)

Answer: \(171\)