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Subtract the sum of $3 l − 4 m − 7 n 2$ and $2 l + 3 m − 4 n 2$ from the sum of $9 l + 2 m - 3 n 2$ and $− 3 l + m + 4 n 2$ .

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l + 4 m + 12 n 2

l + 4 m − 12 n 2

l − 4 m + 12 n 2

l − 4 m − 12 n 2

answer is A.

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Detailed Solution

Given terms are

3 l − 4 m − 7 n 2

2 l + 3 m − 4 n 2

9 l + 2 m - 3 n 2

and

− 3 l + m + 4 n 2

.
According to the question, add

3 l − 4 m − 7 n 2

and

2 l + 3 m − 4 n 2

Now, add

9 l + 2 m - 3 n 2

and

− 3 l + m + 4 n 2

9 l + 2 m − 3 n 2 + − 3 l + m + 4 n 2 = 9 l + 2 m − 3 n 2 − 3 l + m + 4 n 2 ⇒ 9 l + 2 m − 3 n 2 + − 3 l + m + 4 n 2 = 9 l − 3 l + 2 m + m + − 3 n 2 + 4 n 2 ⇒ 9 l + 2 m − 3 n 2 + − 3 l + m + 4 n 2 = 6 l + 3 m + n 2

Subtracting the sum of

3 l − 4 m − 7 n 2

and

2 l + 3 m − 4 n 2

from the sum of

9 l + 2 m - 3 n 2

and

− 3 l + m + 4 n 2

6 l + 3 m + n 2 − 5 l − m − 11 n 2 = 6 l + 3 m + n 2 − 5 l + m + 11 n 2 ⇒ 6 l + 3 m + n 2 − 5 l − m − 11 n 2 = 6 l − 5 l + 3 m + m + n 2 + 11 n 2 ⇒ 6 l + 3 m + n 2 − 5 l − m − 11 n 2 = l + 4 m + 12 n 2

Therefore, the required difference is

l + 4 m + 12 n 2

.
Hence, option 1 is correct.

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