The compound (X) on
(i) on heating in the presence of anhydrous AlCl$$_3$$ and HCl gas gives 2,4-dimethyl pentane
(ii) aromatization gives toluene and
(iii) cyclisation gives methyl cyclohexane 
The correct name of compound (X) is :

First, note that all three reactions mentioned in the statement — isomerisation with $$\text{anh.}\,AlCl_3/HCl$$, aromatisation and cyclisation — are characteristic transformations shown by straight-chain alkanes when treated with suitable catalysts at elevated temperature.

Let the unknown hydrocarbon be $$\mathrm{C_7H_{14}}$$ (an alkene or triene) or $$\mathrm{C_7H_{16}}$$ (an alkane). 

We now match the products obtained in each step with the known behaviour of C-7 hydrocarbons.

Step (1) Isomerisation:

Straight-chain alkanes in the presence of $$\text{anh.}\,AlCl_3$$ and dry $$HCl$$ undergo skeletal rearrangement to give more highly branched isomers having the same molecular formula. For the product to be $$2,4\text{-dimethylpentane}$$ (molecular formula $$\mathrm{C_7H_{16}}$$), the starting compound must also possess the same formula $$\mathrm{C_7H_{16}}$$. The only option given that matches this requirement is heptane (the straight-chain isomer).

Step (2) Aromatisation:

A normal C-7 alkane on dehydrogenation and ring-closure over a Pt/Cr2O3 catalyst forms a C-6 aromatic ring plus one carbon as a side chain, i.e. $$\mathrm{C_6H_5CH_3}$$ (toluene) $$+$$ $$4H_2$$. This is exactly the product stated, confirming that the substrate is the normal (unbranched) C-7 alkane.

Step (3) Cyclisation:

Normal alkanes containing 6-10 carbons cyclise over acidic catalysts (e.g.\ alumina) to give n-1 membered cycloalkanes carrying one alkyl substituent. Thus an n-heptane chain folds to give $$\mathrm{CH_3-(CH_2)_5-CH_3} \longrightarrow \mathrm{CH_3-C_6H_{11}}$$ which is methyl cyclohexane, the product mentioned.

All three observations are therefore satisfied only by the straight-chain alkane $$\mathbf{n\text{-}heptane}$$.

Hence, the correct name of compound (X) is Heptane.

Option C $$\longrightarrow\ $$Heptane