What is the correct IUPAC name of
 

To determine the correct IUPAC name, we need to follow the IUPAC nomenclature rules for substituted benzoic acids.
1. Identify the parent compound: The parent compound is benzoic acid, as it contains a benzene ring with a -COOH group. The -COOH group is assigned position 1.
2. Number the benzene ring: The numbering of the ring is done such that the substituent with the next priority gets the lowest possible number. The priority order for the substituents present is: -COOH (highest priority, assigned position 1)>-OH>-Br>-NO\( _2 \). Therefore, the -OH group should get the lowest possible number. Numbering clockwise gives -OH at position 2, -Br at position 3, and -NO\( _2 \) at position 5. Numbering counterclockwise would give -OH at position 6, which is higher. Thus, the clockwise numbering is correct.
3. Identify the substituents and their positions: - -Br (bromo) is at position 3. - -OH (hydroxy) is at position 2. - -NO\( _2 \) (nitro) is at position 5.
4. Arrange the substituents alphabetically: The alphabetical order of the prefixes is bromo, hydroxy, nitro.
5. Write the IUPAC name: Combining the position numbers and the alphabetically ordered prefixes with the parent name benzoic acid, we get: 3-bromo-2-hydroxy-5-nitrobenzoic acid. This matches option (1).
Let \( y^2 = 12x \) be the parabola and \( S \) its focus. Let \( PQ \) be a focal chord of the parabola such that \( (SP)(SQ) = \frac{147}{4} \). Let \( C \) be the circle described by taking \( PQ \) as a diameter. If the equation of the circle \( C \) is: \[ 64x^2 + 64y^2 - \alpha x - 64\sqrt{3}y = \beta, \] then \( \beta - \alpha \) is equal to:
Let one focus of the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ be at $ (\sqrt{10}, 0) $, and the corresponding directrix be $ x = \frac{\sqrt{10}}{2} $. If $ e $ and $ l $ are the eccentricity and the latus rectum respectively, then $ 9(e^2 + l) $ is equal to:
Let $ A \in \mathbb{R} $ be a matrix of order 3x3 such that $$ \det(A) = -4 \quad \text{and} \quad A + I = \left[ \begin{array}{ccc} 1 & 1 & 1 \\2 & 0 & 1 \\4 & 1 & 2 \end{array} \right] $$ where $ I $ is the identity matrix of order 3. If $ \det( (A + I) \cdot \text{adj}(A + I)) $ is $ 2^m $, then $ m $ is equal to: