The position vectors of two 1 kg particles, (A) and (B), are given by $$ \overrightarrow{r_A}=\left( \alpha_1 t^{2}\widehat{i}+ \alpha_2 t\widehat{j}+\alpha_3 t\widehat{k} \right)m \text{ and } \overrightarrow{r_B}=\left( \beta_1 t\widehat{i} + \beta_2 t^{2}\widehat{j} + \beta_3 t \widehat{k} \right)m,$$  respectively; $$ \left( \alpha_1= 1 m/s^{2}, \alpha_2 = 3nm/s, \alpha_3= 2m/s, \beta_1 = 2m/s, \beta_2 = -1 m/s^{2}, \beta_3 = 4pm/s \right),$$ where t is time, n and p are constants. At $$ t=1s, \mid \overrightarrow{V_A}\mid = \mid \overrightarrow{V_B} \mid \text{ and velocities } \overrightarrow{V_A} \text{ and } \overrightarrow{V_B}\text{ of the particles are orthogonal to each other. At } t=1 s,$$ the magnitude of angular momentum of particle (A) with respect to the position of particle (B) is $$ \sqrt{L} kgm^{2} s^{-1}.\text{ The value of L is }$$ _______ .

We need to find the magnitude of angular momentum of particle A with respect to the position of particle B at t = 1s.

$$\vec{r_A} = (\alpha_1 t^2 \hat{i} + \alpha_2 t \hat{j} + \alpha_3 t \hat{k})$$ m

$$\vec{r_B} = (\beta_1 t \hat{i} + \beta_2 t^2 \hat{j} + \beta_3 t \hat{k})$$ m

$$\alpha_1 = 1, \alpha_2 = 3n, \alpha_3 = 2, \beta_1 = 2, \beta_2 = -1, \beta_3 = 4p$$

$$\vec{V_A} = \frac{d\vec{r_A}}{dt} = (2\alpha_1 t \hat{i} + \alpha_2 \hat{j} + \alpha_3 \hat{k})$$

At t = 1: $$\vec{V_A} = (2 \hat{i} + 3n \hat{j} + 2 \hat{k})$$ m/s

$$\vec{V_B} = \frac{d\vec{r_B}}{dt} = (\beta_1 \hat{i} + 2\beta_2 t \hat{j} + \beta_3 \hat{k})$$

At t = 1: $$\vec{V_B} = (2 \hat{i} - 2 \hat{j} + 4p \hat{k})$$ m/s

$$4 + 9n^2 + 4 = 4 + 4 + 16p^2$$

$$9n^2 + 4 = 4 + 16p^2$$

$$9n^2 = 16p^2$$ ... (i)

$$\vec{V_A} \cdot \vec{V_B} = 0$$

$$2(2) + 3n(-2) + 2(4p) = 0$$

$$4 - 6n + 8p = 0$$

$$6n - 8p = 4$$

$$3n - 4p = 2$$ ... (ii)

From (i): $$3n = 4p$$ (taking positive root, but let's check both)

If $$3n = 4p$$, substituting in (ii): $$4p - 4p = 2 \Rightarrow 0 = 2$$ (contradiction)

If $$3n = -4p$$, substituting in (ii): $$-4p - 4p = 2 \Rightarrow -8p = 2 \Rightarrow p = -1/4$$

Then $$3n = -4(-1/4) = 1 \Rightarrow n = 1/3$$

$$\vec{r_A}(1) = (1 \hat{i} + 3(1/3) \hat{j} + 2 \hat{k}) = (1 \hat{i} + 1 \hat{j} + 2 \hat{k})$$ m

$$\vec{r_B}(1) = (2 \hat{i} + (-1) \hat{j} + 4(-1/4) \hat{k}) = (2 \hat{i} - 1 \hat{j} - 1 \hat{k})$$ m

$$\vec{V_A}(1) = (2 \hat{i} + 1 \hat{j} + 2 \hat{k})$$ m/s

$$\vec{r} = \vec{r_A} - \vec{r_B} = (1-2)\hat{i} + (1+1)\hat{j} + (2+1)\hat{k} = (-1\hat{i} + 2\hat{j} + 3\hat{k})$$ m

$$\vec{L} = m(\vec{r} \times \vec{V_A}) = 1 \times \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & 3 \\ 2 & 1 & 2 \end{vmatrix}$$

$$= \hat{i}(2 \times 2 - 3 \times 1) - \hat{j}((-1)(2) - 3(2)) + \hat{k}((-1)(1) - 2(2))$$

$$= \hat{i}(4-3) - \hat{j}(-2-6) + \hat{k}(-1-4)$$

$$= \hat{i}(1) - \hat{j}(-8) + \hat{k}(-5)$$

$$= (1\hat{i} + 8\hat{j} - 5\hat{k})$$

$$|\vec{L}| = \sqrt{1 + 64 + 25} = \sqrt{90}$$

Since $$|\vec{L}| = \sqrt{L}$$, we get $$L = 90$$.

The answer is 90.