\( (\vec{a} + \vec{b}) \times \vec{c} - \vec{c} \times (-2\vec{a} + 3\vec{b}) = 0 \)
\( (\vec{a} + \vec{b}) \times \vec{c} + (-2\vec{a} + 3\vec{b}) \times \vec{c} = 0 \)
\( \Rightarrow (\vec{a} + \vec{b}) - 2\vec{a} + 3\vec{b} \times \vec{c} = 0 \)
\( \Rightarrow \vec{c} = \lambda (\vec{4b} - \vec{a}) \)
\( \Rightarrow \lambda(44\hat{i} - 4\hat{j} + 4\hat{k} - 4\hat{i} + \hat{j} - \hat{k}) \)
\( = \lambda(40\hat{i} - 3\hat{j} + 3\hat{k}) \)
Now
\( (8\hat{i} - 2\hat{j} + 2\hat{k} + 33\hat{i} - 3\hat{j} + 3\hat{k}) \cdot \lambda(40\hat{i} - 3\hat{j} + 3\hat{k}) = 1670 \)
\( \Rightarrow (41\hat{i} - 5\hat{j} + 5\hat{k}) \cdot (40\hat{i} - 3\hat{j} + 3\hat{k}) = \lambda 1670 \)
\( \Rightarrow (1640 + 15)\lambda = 1670 \Rightarrow \lambda = 1 \)
So \( \vec{c} = 40\hat{i} - 3\hat{j} - 3\hat{k} \)
\( \Rightarrow |\vec{c}|^2 = 1600 + 9 + 9 = 1618 \)
Given: \[ (\vec{a} + \vec{b}) \times \vec{c} = \vec{c} \times (-2\vec{a} + 3\vec{b}). \]
Expanding: \[ (\vec{a} + \vec{b}) \times \vec{c} + \vec{c} \times (-2\vec{a} + 3\vec{b}) = 0. \]
This implies that: \[ \vec{c} = \lambda (4\vec{b} - \vec{a}). \]
We substitute the values of \(\vec{a}\) and \(\vec{b}\): \[ \vec{c} = \lambda \left( 4(11\hat{i} - \hat{j} + \hat{k}) - (4\hat{i} - \hat{j} + \hat{k}) \right), \] \[ \vec{c} = \lambda (40\hat{i} - 3\hat{j} + 3\hat{k}). \]
Given that: \[ (2\vec{a} + 3\vec{b}) \cdot \vec{c} = 1670. \]
Calculating \( 2\vec{a} + 3\vec{b} \): \[ 2\vec{a} + 3\vec{b} = 2(4\hat{i} - \hat{j} + \hat{k}) + 3(11\hat{i} - \hat{j} + \hat{k}), \] \[ 2\vec{a} + 3\vec{b} = (8 + 33)\hat{i} + (-2 - 3)\hat{j} + (2 + 3)\hat{k}, \] \[ 2\vec{a} + 3\vec{b} = 41\hat{i} - 5\hat{j} + 5\hat{k}. \]
Now, we compute: \[ (2\vec{a} + 3\vec{b}) \cdot \vec{c} = (41\hat{i} - 5\hat{j} + 5\hat{k}) \cdot \lambda (40\hat{i} - 3\hat{j} + 3\hat{k}), \] \[ (2\vec{a} + 3\vec{b}) \cdot \vec{c} = \lambda (41 \cdot 40 + (-5) \cdot (-3) + 5 \cdot 3), \] \[ 1670 = \lambda (1640 + 15 + 15), \] \[ 1670 = 1670\lambda \implies \lambda = 1. \]
Thus: \[ \vec{c} = 40\hat{i} - 3\hat{j} + 3\hat{k}. \]
Calculating \( |\vec{c}|^2 \): \[ |\vec{c}|^2 = 40^2 + (-3)^2 + 3^2, \] \[ |\vec{c}|^2 = 1600 + 9 + 9, \] \[ |\vec{c}|^2 = 1618. \]
Therefore: \[ 1618. \]
Consider the following sequence of reactions : 
Molar mass of the product formed (A) is ______ g mol\(^{-1}\).
In a Young's double slit experiment, three polarizers are kept as shown in the figure. The transmission axes of \( P_1 \) and \( P_2 \) are orthogonal to each other. The polarizer \( P_3 \) covers both the slits with its transmission axis at \( 45^\circ \) to those of \( P_1 \) and \( P_2 \). An unpolarized light of wavelength \( \lambda \) and intensity \( I_0 \) is incident on \( P_1 \) and \( P_2 \). The intensity at a point after \( P_3 \), where the path difference between the light waves from \( S_1 \) and \( S_2 \) is \( \frac{\lambda}{3} \), is:
