Show that the lines `(x+1)/(-3)=(y-3)/(2)=(z+2)/(1)and(x)/(1)=(y-7)/(-3)=(z+7)/(2)` are coplanar. Also, find the equation of the plane containing them.

Explanation

Here , `a_(1)=-1,b_(1)=3,c_(1)=-2`
`a_(2)=0,b_(2)=7,c_(2)=-7`
`l_(1)=-3,m_(1)=2,n_(1)=1`
`l_(2)=1,m_(2)=-3,n_(2)=2`
`[{:(a_(2)-a_(1),b_(2)-b_(1),c_(2)-c_(1)),(l_(1),m_(1),n_(1)),(l_(2),m_(2),n_(2)):}]=[{:(1,4,-5),(-3,2,1),(1,-3,2):}]`
`=1(4+3)-4(-6-1)-5(9-2)`
`=7+28-35=0`
Hence,lines are coplanar.
Plane is passing through the point (-1,3,-2) and (0,7,-7). Equation of plane passing through the point (-1,3,-2),
`a(x+1)+b(y-3)+c(z+2)=0` ...(i)
Equation pf plane passing through the point (0,7,-7),a+4b-5c=0 ...(ii)
Plane is perpendicular to (-3,2,1),
`therefore-3a+2b+c=0` ...(iii)
By equations (ii) and (iii) , we get
`(a)/(14)=(b)/(15-1)=(c)/(2+12)=lambda`(let)
`(a)/(1)=(b)/(1)=(c)/(1)=lambda`
`a=b=c=lambda`
Put the value a ,b and c in equation (i)
`lambda(x+1)+lambda(y-3)+lambda(z+2)=0`
`x+y+z+3-3=0`
`x+y+z=0`