The correct answer is 16
As slope of line joining (1, 2) and (3, 6) is 2 given diameter is parallel to side
Hence, \(a = \sqrt{(3−1)^2+(6−2)^2}\)
\(= \sqrt{20}\)
Thereafter, \(\frac{b}{2} = \frac{4}{\sqrt5}\)
\(⇒ b = \frac{8}{\sqrt5}\)
So, the Area ab \(= 2 \sqrt5 . \frac{8}{\sqrt5}\)
= 16
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} $
Resonance in X$_2$Y can be represented as
The enthalpy of formation of X$_2$Y is 80 kJ mol$^{-1}$, and the magnitude of resonance energy of X$_2$Y is: