The relation between heat at constant pressure (\( \Delta H \)) and at constant volume (\( \Delta U \)) is:
\( \Delta H = \Delta U + \Delta n_g RT \)
For benzoic acid:
\( C_6H_5COOH(s) + \frac{15}{2} O_2(g) \rightarrow 7CO_2(g) + 3H_2O(l) \)
\( \Delta n_g = 7 - \frac{15}{2} = -\frac{1}{2} \). Substituting:
\( \Delta H = -321.30 - \frac{1}{2} R \times 300 \)
Here, \( R \approx 8.314 \, \text{J/mol.K} \). Solving gives:
\( x = 150 \)
Let a line passing through the point $ (4,1,0) $ intersect the line $ L_1: \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} $ at the point $ A(\alpha, \beta, \gamma) $ and the line $ L_2: x - 6 = y = -z + 4 $ at the point $ B(a, b, c) $. Then $ \begin{vmatrix} 1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{vmatrix} \text{ is equal to} $