Correct Answer - C
We have,
`vec(a),vec(b)` and `vec(c)` are mutually perpendicular vectors.
`:.vec(a).vec(b)=vec(b).vec(c)=vec(a).vec(c)=0`
and `|vec(a)|=1,|vec(b)|=2, |vec(c)|=3`
Now, `[vec(a)+vec(b)+vec(c) vec(b)-vec(a)vec(c)]`
`=(vec(a)+vec(b)+vec(c))*((vec(b)-vec(a))xx vec(c))`
`=(vec(a)+vec(b)+vec(c))*[(vec(b)xx vec(c))-(vec(a)xx vec(c))]`
`=vec(a)*( vec(b) xx vec(c))-vec(a).(vec(a)xx vec(c))+ vec(b)* (vec(b)xx vec(c))-vec(b)*(vec(a)xx vec(c))+ vec(c)* (vec(b)xx vec(c))-vec(c)* (vec(a)xx vec(c))`
`=[vec(a) vec(b) vec(c)]- [vec(a) vec(a) vec(c)] + [ vec(b)vec(b)vec(c)]- [ vec(b) vec(a) vec(c)] + [ vec(c) vec(b) vec(c)]- [ vec(c)vec(a)vec(c)]`
`= [ vec(a)vec(b)vec(c)]-0+0+[vec(a)vec(b)vec(c)]+0-0`
`=2 [ vec(a) vec(b) vec(c)]`
`= 2vec(a)* ( vec(b)xx vec(c))=2vec(a)*(|vec(b)||vec(c)| sin. (pi)/(2) n)`
`=2vec(a) *(2xx3xx1xx hat(n))=12vec(a)* hat(n)`
`= 12 | vec(a)||hat(n)| cos 0^(@)=12xx1xx1xx1=12`
Hence, option (c) is correct.