We start with the given expression:
\[ |3 \, \text{adj}(2 \, \text{adj}(|A|A))| \]
Now simplify step-by-step:
\[ |3 \, \text{adj}(2|A|^2 \, \text{adj}A)| \]
\[ |3 \cdot 2^2 \cdot |A|^4 \, \text{adj}(\text{adj}A)| \]
We know that \(|\text{adj}A| = |A|^{n-1}\). Hence, substituting and simplifying:
\[ 3^3 \cdot 2^6 \cdot |A|^{12} \cdot |A|^2 = 3^{-13} \cdot 2^{-10} \]
Now,
\[ |A|^{16} = 3^{-16} \cdot 2^{-16} \]
\[ |A| = \frac{1}{6} \]
Substitute back into the main equation:
\[ |3 \, \text{adj}(2A)| = |3 \cdot 2^2 \, \text{adj}A| = 3^3 \cdot 2^6 |A|^2 \]
\[ = 3 \cdot 2^4 \]
Therefore,
\[ M = 4, \quad n = 1 \]
And,
\[ |3m + 2n| = 14 \]
Step 1: General equation of the circle The equation of the circle is:
\[ (x - h)^2 + (y - k)^2 = r^2. \]
Since the circle passes through the points \((0, 0)\) and \((1, 0)\), substitute these points into the circle’s equation.
For \((0, 0)\):
\[ h^2 + k^2 = r^2. \]
For \((1, 0)\):
\[ (1 - h)^2 + k^2 = r^2. \]
Expanding and simplifying:
\[ 1 - 2h + h^2 + k^2 = h^2 + k^2. \]
Substitute \(r^2 = h^2 + k^2\) into the equation:
\[ 1 - 2h + h^2 + k^2 = h^2 + k^2. \]
Cancel \(h^2 + k^2\):
\[ 1 - 2h = 0 \implies h = \frac{1}{2}. \]
Step 2: Circle touches \(x^2 + y^2 = 9\) The given circle \(x^2 + y^2 = 9\) has a radius \(R = 3\) and is centered at \((0, 0)\). For the circle to touch \(x^2 + y^2 = 9\), the distance between their centers must be equal to the difference of their radii:
\[ \sqrt{h^2 + k^2} = R - r = 3 - \sqrt{h^2 + k^2}. \]
Let \(d = \sqrt{h^2 + k^2}\):
\[ d = 3 - d \implies 2d = 3 \implies d = \frac{3}{2}. \]
Thus:
\[ h^2 + k^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4}. \]
Step 3: Compute \(4(h^2 + k^2)\) Multiply \(h^2 + k^2\) by 4:
\[ 4(h^2 + k^2) = 4 \cdot \frac{9}{4} = 9. \]
Final Answer: 9.

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:
