The correct option is C −42
Given: →b=2^i+^j+^k,→c=^i−^j+^k
→d=3^i+2^j+6^k
|→a|=√10
Now, [→b →c →d]=∣∣
∣∣2111−11326∣∣
∣∣
=2(−6−2)−1(3)+1(5)=−14
As →a is coplanar with vectors →b and →c, so assuming
→a=λ→b+μ→c⇒→a=(2λ+μ)^i+(λ−μ)^j+(λ+μ)^k
We know, →a is perpendicular to →d
⇒→a⋅→d=0⇒3(2λ+μ)+2(λ−μ)+6(λ+μ)=0⇒14λ++7μ=0⇒μ=−2λ ⋯(1)
Also, |→a|=√10
⇒(2λ+μ)2+(λ−μ)2+(λ+μ)2=10
From equation (1), we get
⇒(2λ−2λ)2+(λ+2λ)2+(λ−2λ)2=10⇒10λ2=10⇒λ=±1⇒μ=∓2
Now,
[→a →b →c]+[→a →b →d]+[→a →c →d]=0+[→a →b →d]+[→a →c →d]=[(λ→b+μ→c) →b →d]+[(λ→b+μ→c) →c →d]=λ[→b →b →d]+μ[→c →b →d]+λ[→b →c →d]+μ[→c →c →d]=−μ[→b →c →d]+λ[→b →c →d]=(λ−μ)[→b →c →d]
From equation (1), we get
=−42λ=∓42