Answer with explanation:
Equation of Plane having Direction cosines A, B and C passing through points, p, q and r is
⇒A (x-p)+B(y-q)+C(z-r)=0
The plane passes through the points A(-2, 3,-1), B(0, 5, 2) and C(-1, -2, 1).
→A(x+2)+B(y-3)+C(z+1)=0----------(1)
→A(0+2)+B(5-3)+C(2+1)=0
2 A +2 B+3 C=0
→A(-1+2)+B(-2-3)+C(1+1)=0
A -5 B+2 C=0
[tex]\Rightarrow \frac{A}{4+15}=\frac{B}{3-4}=\frac{C}{-10-2}\\\\\Rightarrow \frac{A}{19}=\frac{B}{-1}=\frac{C}{-12}=k\\\\A=19 K,B=-K, C=-12K[/tex]
Substituting the value of A , B and C in equation (1)
⇒19 K(x+2)-K(y-3)-12 K(z+1)=0
⇒19 x +38 -y +3-12 z-12=0
⇒19 x -y -12 z +29=0, is the required equation of Plane in Cartesian form.
⇒(19 i -j -12 k)(xi +y j+z k)+29=0 ,is the required vector equation of the plane.
