If →a and →b are vectors such that |→a+→b|=√29 and →a×(2^i+3^j+4^k)=(2^i+3^j+4^k)×→b, then a possible value of (→a+→b)⋅(−^i+2^j+3^k) is

→a×(2^i+3^j+4^k)=(2^i+3^j+4^k)×→b
→a×(2^i+3^j+4^k)=−→b×(2^i+3^j+4^k)
→a×(2^i+3^j+4^k)+→b×(2^i+3^j+4^k)=0
(→a+→b)×(2^i+3^j+4^k)=0
This implies that (→a+→b) is parallel to (2^i+3^j+4^k).
Unit vector in the direction of →a+→b is ±(2^i+3^j+4^k)/√29
Also given that ||→a+→b||=√29
∴→a+→b=√29×[±(2^i+3^j+4^k)/√29]=±(2^i+3^j+4^k)
Hence (→a+→b)⋅(−^i+2^j+3^k)=±(2^i+3^j+4^k)⋅(−^i+2^j+3^k)
=±(−2+6+12)=±16
However, the question has −^i+2^j+3^k instead of −^i+2^j+2^k. Let's assume it's a typo and it should be (−^i+2^j+2^k).
Then (→a+→b)⋅(−^i+2^j+2^k)=±(2^i+3^j+4^k)⋅(−^i+2^j+2^k)
=±(−2+6+8)=±12
If the vector is (−^i+2^j+3^k), then
(→a+→b)⋅(−^i+2^j+3^k) = ±(2(−1)+3(2)+4(3)) = ±( -2+6+12) = ±16
There must be some error in the question or options provided. Let's re-examine the given condition:
(→a+→b)×(2i+3j+4k) = 0
This means that (→a+→b) is parallel to (2i+3j+4k).
Since |→a+→b| = √29, we have →a+→b = ±(2i+3j+4k)
Then (→a+→b)⋅(−i+2j+3k) = ±(2(−1)+3(2)+4(3)) = ±16
Therefore, a possible value is 16 (or -16). However, 16 is not among the options. There is likely a mistake in the question or options.