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I proved this by induction. But I want to show it using modular arithmetic. I tried for sometime as follows

$$2×7^n-2+3×5^n-3\
2(7^n-1)+3(5^n-1)\
2×6a+3×4b\
12(a+b)$$

In this way I just proved that it is divisible by 12 but it is not enough. Am I missing something or it will solved by some other method.

Yes, it can be done by another method. Note that $7^2=2times24+1$ and that $5^2=24+1$ and that therefore$$7^nequivbegincases7pmod24&text if $n$ is odd\1pmod24&text otherwiseendcases$$and$$5^nequivbegincases5pmod24&text if $n$ is odd\1pmod24&text otherwise.endcases$$So:

• if $n$ is odd, then $2times7^n+3times5^n-5equiv2times7+3times5-5=24equiv0pmod24$;


• otherwise, $2times7^n+3times5^n-5equiv2times1+3times1-5equiv0pmod24$.

$$2(7^n-1)+3(5^n-1)$$

$$=2((1+6)^n-1)+3((1+4)^n-1)$$

$$equiv2(6n+text terms containing 6^2)+3(4n+text terms containing 4^2)$$

$$equiv 24npmod24$$

Case 1 : $n$ is odd

In this case $$2×7^n+3×5^n-5=2×7^n+2×5^n+5^n-5\=2timesunderbrace(7^n+5^n)_12k+5(5^n-1-1)\=24k+5(underbrace25^n-1over 2-1_24k')\=24k''$$

Case 2 : $n$ is even

In this case $$2×7^n+3×5^n-5=14×7^n-1+15×5^n-1-5\=14timesunderbrace(7^n-1+5^n-1)_12k+5(5^n-2-1)\=24k+5(underbrace25^n-2over 2-1_24k')\=24k''$$

Note that you have
$$
7^n - 1 = 6a\
5^n - 1 = 4b
$$
Now we're interested in whether $a$ and $b$ are even or odd. Which is to say we want to know when $7^n - 1$ is divisible by $4$ (so that when you divide it by $6$ you get an even number), and when $5^n-1$ is divisible by $8$ (so that when you divide it by $4$, you get an even number).

The binomial theorem gives
$$
7^n - 1 = (8-1)^n - 1\
= 8^n - binom n18^n-1 + cdots + (-1)^n-1binomnn-18 + (-1)^n - 1
$$
We see that this is divisible by $4$ exactly when $(-1)^n - 1$ is, which is to say when $n$ is even.

Then we have
$$
5^n - 1 = (4 + 1)^n - 1\
= 4^n + binom n14^n-1 + cdots + binomnn-14 + 1 - 1
$$
and we see that this is divisible by $8$ precisely when $binomnn-1 = n$ is even.

So $a$ and $b$ are both even for even $n$, and both odd for odd $n$, proving that $a+b$ is always even, meaning $12(a+b)$ is divisible by $24$.

You may split it up by calculating $mod 8$ and $mod 3$:

• 
  $mod 8$:
  begineqnarray* 2×7^n+3×5^n-5
  & equiv_8 & 2times (-1)^n + 3times (-3)^n +3 \
  & equiv_8 & 2times (-1)^n + 3((-3)^n + 1)\
  & stackrel3^2 equiv_8 1equiv_8& begincases 2+3times (1+1) & n = 2k \ -2 +3 (-3 + 1) & n= 2k+1endcases\
  & equiv_8 & 0
  endeqnarray*
  


• 
  $mod 3$:
  begineqnarray* 2×7^n+3×5^n-5
  & equiv_3 & 2times 1^n + 3times (-1)^n +1 \
  & equiv_3 & 3times (1 + (-1)^n)\
  & equiv_3 & 0
  endeqnarray*
  

$$2×7^n-2+3×5^n-3\
2(7^n-1)+3(5^n-1)\
2×6a + 3×4b\
12(color#c00a+b) $$
but it is not enough [to prove divisibility by $24$]

Finish simply with $ 2mid color#c00a+b, =, dfrac7^large n!-17-1 + dfrac5^large n!-15-1, =, overbrace7^large n-1!+5^large n-1^rm even +cdots + overbrace7+5^rm even, +, overbrace1+1^rm even$