Correct Answer - A
Let `r=xhat(i)+yhat(j)+zhat(k)`
Given `rxxb=cxxb`
`rArr (xhat(i)+yhat(j)+zhat(j))xx(hat(i)+hat(j)+hat(k))`
`=(4hat(i)-3hat(j)+7hat(k))xx(hat(i)+hat(j)+hat(k))`
`rArr (y-z)hat(i)-(x-z)hat(j)+(x-y)hat(k)=-10hat(i)+3hat(j)+7hat(k)`
`because|{:(hat(i),hat(j),hat(j)),(x,y,z),(1,1,1):}|=hat(i)(y-z)-hat(j)(x-z)+hat(k)(y-z)`
`rArr|{:(hat(i),hat(j),hat(j)),(4,-3,7),(1,1,1):}|=hat(i)(-10)-hat(j)(-3)+hat(k)(7)`
`rArr y-z=-10,-(x-z)=3,x-y=7`
`rArr y-z=-10,-x+z=3,x-y=7`....(i)
and `r.a=0`
`rArr (xhat(i)+yhat(j)+zhat(k)).(2hat(i)+hat(k))`
`rArr 2x+z=0` ......(ii)
From Eqs.(i) and (ii) , we get
`x=-1,y=-8,z=2`
`therefore r.b=(-hat(i)-8hat(j)+2hat(k)).(hat(i)+hat(j)+hat(k))`
`=-1-8+2=-7`.