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 (Lorentz) force on a moving point charge is given by
$$\vec{F}=q\,\bigl(\vec{v}\times\vec{B}\bigr)$$

Given data:
Charge, $$q = 1\;\mu\text{C} = 1\times10^{-6}\,\text{C}$$
Velocity, $$\vec{v}=1\,\hat{i}-2\,\hat{j}+3\,\hat{k}\;(\text{m s}^{-1})$$
Magnetic field, $$\vec{B}=2\,\hat{i}+3\,\hat{j}-5\,\hat{k}\;(\text{T})$$

First, calculate the cross product $$\vec{v}\times\vec{B}$$.

Using the determinant form:
$$ \vec{v}\times\vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 1 & -2 & 3\\ 2 & 3 & -5 \end{vmatrix} $$

Expand the determinant:

$$ \vec{v}\times\vec{B} = \hat{i}\bigl((-2)(-5) - (3)(3)\bigr) - \hat{j}\bigl(1(-5) - 3(2)\bigr) + \hat{k}\bigl(1\cdot3 - (-2)\cdot2\bigr) $$
$$ = \hat{i}\bigl(10 - 9\bigr) - \hat{j}\bigl(-5 - 6\bigr) + \hat{k}\bigl(3 + 4\bigr) $$
$$ = 1\,\hat{i} + 11\,\hat{j} + 7\,\hat{k} $$

Magnitude of the cross product:
$$ \lvert\vec{v}\times\vec{B}\rvert = \sqrt{1^{2}+11^{2}+7^{2}} = \sqrt{1+121+49} = \sqrt{171} $$

Therefore, the magnitude of the magnetic force is
$$ F = q\,\lvert\vec{v}\times\vec{B}\rvert = \bigl(1\times10^{-6}\bigr)\sqrt{171}\;\text{N} = \sqrt{171}\times10^{-6}\;\text{N} $$

Comparing with the required form $$\sqrt{\alpha}\times10^{-6}\,\text{N}$$, we have $$\alpha = 171$$.

Hence, the value of $$\alpha$$ is 171.