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Q6. Sum of a, b, c and d is $120$, such that their HCF is $3$, How many solutions are possible ?

I have tried this ...

I converted the above scene into the following equation as

$ 3a+3b+3c+3d=120$

so a+b+c+d=40

so integral solutions to the equations are ${40 +4-1\choose 4-1}$

Is it correct?

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    No -- you need to correct the count to remove quadruples $(a,b,c,d)$ which have a common factor greater than $1$. – 2017-01-19
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    You appear to be ignoring the $\gcd$ condition. after you divide by $3$ you need the four integers to be relatively prime. Also you should say if you mean the pairwise $\gcd$ or the $\gcd$ of the four collectively. – 2017-01-19
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    Also, what were the stated restrictions on $a,b,c,d$? Positive integers? Nonnegative integers? – 2017-01-19
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    +ve integers @quasi – 2017-01-19
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    Then you were using the wrong formula. – 2017-01-19
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    Also, are you familiar with "PIE" (the Principle of Inclusion-Exclusion)? – 2017-01-19

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$\newcommand{'}{\ \!'}$ That's a good start, and it tacitly assumes that zero is a permitted value of the components, but you need to also remove solutions with higher common factors than $3$ in the sum to $120$.

So you already have $a\'+b\'+c\'+d\' = 40$ - you need to remove the cases where $\gcd(a\',b\',c\',d\') = 2 $ or $5$ and then add back in the cases where $\gcd(a\',b\',c\',d\') = 10$ (which would be "twice-removed"). This is a typical inclusion-exclusion process.

If zero is not permitted, you will need to modify your initial count to $\binom {40-1}{4-1}$ to allow for each variable to be a positive value, and similarly with the subsequent steps.