Let Q be the image of the point P(i+3k) in the plane r⋅(i+j+k)=1. Then, PQ is normal to the plane. Since, PQ passes through P and in normal to the given plane, therefore equation of PQ is r=(i+3k)+λ(i+j+k)
Since, Q lies on the line PQ, so let the position vector of Q be (i+3k)+λ(i+j+k)
i.e. (1+λ)i+λj+(3+λ)k
Since, R is the mid-point of PQ, therefore position vector of R is 2(1+λ)i+λj+(3+λ)k+i+3k =(2λ+2)i+2λj+(26+λ)k =(2λ+1)i+2λj+(3+2λ)k
Since, R lies on the plane r⋅(i+j+k)=1 ∴[(2λ+1)i+2λj+(3+2λ)k] ⋅(i+j+k)=1 ⇒[2λ+1+2λ+3+2λ]=1 ⇒λ=−2
Hence, the position vector of Q is −i−2j+k.