∴DR of PQ=DR of line =⟨−2,2,1⟩ ∴ Equation of line PQ passing through P(3,2,−1) and DR=<−2,2,1> is
x−3
−2
=
y−2
2
=
z+1
1
Any General point on line PQ=(x1,y1,z1) ∴
x1−3
−2
=
y1−2
2
=
z1+1
1
=λ ⇒x1=−2λ+3 y1=2λ+2 z1=λ−1 ∴ Point Q=(−2λ+3,2λ+2,λ−1) Point Q lies on the plane 3x−y+4z+1=0. So point Q satisfy the equation. 3(−2λ+3)−(2λ+2)+4(λ−1)+1=0 ⇒−6λ+9−2λ−2+4λ−4+1=0 ⇒−4λ+4=0 ⇒λ=1 ∴ Point Q =(−2×1+3,2×1+2,1−1) =(1,4,0) ∴ Distance of the point P(3,2,−1) from the plane = Length of PQ =√(3−1)2+(2−4)2+(−1−0)2 =√22+(−2)2+(−1)2 =√4+4+1 =√9 =3