There are $$3$$ pineapples, $$6$$ bananas and $$7$$ apples. Each fruit of the same category has the same price. The total amount of the fruits in the first row is Rs. $$44$$, that of the third row is Rs. $$54$$ and that of the first column is Rs. $$72$$. Then the total amount in rupees of the fruits in the second row is

Let the cost of one apple be $$a$$, one banana be $$b$$, and one pineapple be $$p$$.

From the first row,

$$a+b=22.$$

From the third row,

$$p+a+2b=54.$$

From the first column,

$$2a+p+b=72.$$

From the first equation,

$$a=22-b.$$

Substitute this into the second equation.

$$p+(22-b)+2b=54.$$

$$p+b=32.$$

So,

$$p=32-b.$$

Now substitute $$a=22-b$$ and $$p=32-b$$ into the third equation.

$$2(22-b)+(32-b)+b=72.$$

$$44-2b+32-b+b=72.$$

$$76-2b=72.$$

$$2b=4.$$

$$b=2.$$

Now,

$$a=22-2=20,$$

and

$$p=32-2=30.$$

The cost of the second row is

$$a+2p+b=20+2(30)+2=82.$$

Hence, the cost of the second row is $$82$$