Given,
→a=2^i+3^j+^k
→b=^i−2^j+^k
→c=−3^i+^j+2^k
The value of [→a→b→c] is given by (→a×→b)⋅→c
→a×→b=⎡⎢⎣^i^j^k2311−21⎤⎥⎦
=(3×1−1(−2))^i−(2×1−1×1)^j+(2(−2)−3×1)^k
=5^i−^j−7^k
Now, (→a×→b)⋅→c
=(5^i−^j−7^k)⋅(−3^i+^j+2^k)
=−15−1−14
=−30
Alternate:
[→a→b→c] is equal to the determinant of the matrix that has the three vectors either as its rows or its columns.
∴[→a→b→c]=∣∣
∣∣2311−21−312∣∣
∣∣
=2(−4−1)−3(2+3)+1(1−6)=−30