Three Dimensional Geometry Part 2

Line PQ is parallel to line ‌

2−x

2

=‌

y−3

2

=‌

z+1

1

∴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