The number is successively divided by 5, 7, 11 give remainder as 3, 1, 10.

If last quotient is 114.
Find the numbers.

3 answers

I will attempt the simplest approach:
numbers which when divided by 5 leave a remainder of 3 :
8, 13, 18, 23, 28, 33, 38, 43, 48, 53, 58, ...

numbers which when divided by 7 leave a remainder of 1
8, 15, 22, 29, 36, 43, 50, 57, 64, ....

numbers which when divided by 11 leave a remainder of 10
10, 21, 32, 43, 54 ...

ahhh, notice 43 satisfies all 3 conditions.

This was lucky, if the number had been larger, this method would become
impractical. In that case I had hinted in another reply that I would use
"The Chinese Remainder Theorem"

btw, if we multiply the divisors, and then add or subtract multiples of
that product we get another number that satisfies.
so 5*7*11 = 385

which means 43+385 or 428 would also work , you can check it !

so possible answers are 43 + k(385) , where k is a whole number.
Sorry dear
Didn't understand
Kindly explain in another way
Please
I don't understand that you "don't understand"
I gave you the simplest and most direct way of doing this.
I repeat my simple way:

I will attempt the simplest approach:
numbers which when divided by 5 leave a remainder of 3 :
8, 13, 18, 23, 28, 33, 38, 43, 48, 53, 58, ...

numbers which when divided by 7 leave a remainder of 1
8, 15, 22, 29, 36, 43, 50, 57, 64, ....

numbers which when divided by 11 leave a remainder of 10
10, 21, 32, 43, 54 ...

ahhh, notice 43 satisfies all 3 conditions.

It said find the numbers, I assumed you wanted more than just the 43.
That's where the last part comes in.