To solve this problem, we need to analyze the structure of pyridine.$\\$Pyridine is a six-membered heterocyclic aromatic compound with the formula $\mathrm{C_5H_5N}. \\$It consists of a benzene-like ring where one of the carbon atoms is replaced by a nitrogen atom.$\\$Let's break down the structure:$\\$• Pyridine has a total of six atoms in the ring: five carbon atoms and one nitrogen atom.$\\$• Each carbon atom forms three sigma bonds: two with adjacent carbon atoms and one with a hydrogen atom.$\\$• The nitrogen atom forms three sigma bonds: two with adjacent carbon atoms and one with its lone pair.$\\$Now, let's count the sigma bonds:$\\$• Each of the five carbon atoms forms three sigma bonds, contributing a total of $5 \times 3 = 15$ sigma bonds.$\\$• However, this count includes the carbon-carbon sigma bonds twice, so we need to adjust for this.$\\$• There are five carbon-carbon sigma bonds in the ring.$\\$• Therefore, the correct count of sigma bonds is:$\\ 15 - 5 = 10$ (from carbon-carbon bonds)$+ 5$ (carbon-hydrogen bonds)$+ 1$ (nitrogen lone pair)$= 11$ sigma bonds.$\\$Next, let's count the pi bonds:$\\$• Pyridine has three carbon-carbon double bonds, each contributing one pi bond.$\\$• Therefore, there are 3 pi bonds in pyridine.$\\$Finally, let's consider the lone pairs:$\\$• The nitrogen atom in pyridine has one lone pair of electrons.$\\$Therefore, the number of sigma bonds, pi bonds, and lone pairs in pyridine are:$\\$11 sigma bonds, 3 pi bonds, and 1 lone pair of electrons.$\\$This corresponds to Option 3: 11, 3, 1.