The vector equation of the plane passing tgrough the intersection of planes `barr.(2hati-3hatj+4hatk)=1and barr.(hati-hatj)+4=0` and perpendicular to the plane `barr.(2bari-hatj+hatk)=-5` is

Explanation

The vector equation of the plane passing through the intersection of the planes
`barr*barn_(1) = p_(1) and barr*barn_(2) = p_(2)` is
`barr*(bar(n_(1))+lambdabar(n_(1)))=p_(1)+lambdap_(2)`, where `lambda` is a parameter.
Here, `bar(n_(1)) = 2hati - 3hatj + 4hatk, bar(n_(2)) = hati - hatj, p_(1) 1` and `p_(2) =-4`
`therefore` the vector equation of the plane passing through the intersection of the given planes is
`barr*[(2hati-3hatj+4hatk)+lambda(hati-hatj)]=1+lambda(-4)`
`therefore barr*[(2+lambda)hati+(-3-lambda)hatj+4hatk]=1-4lambda" "...(1)`
Since teh plane is perpendicular to the plane
`barr*(2hati-hatj+hatk)=-5,` their normal vectors
`(2+ lambda) hati + (-3-lambda)hatj + 4hatk` and `2hati - hatj + hatk` are perpendicular.
`therefore [(2+lamda)hati+(-3-lambda)hatj+4hatk]*(2hati-hatj+hatk)=0`
`therefore (2+lambda)(2)+(-3-lambda)(-1)+4(1)=0`
`therefore 4 + 2lambda + 3 + lambda + 4 = 0`
`therefore 3 lambda + 11 = 0 " " therefore lambda = - (11)/(3)`
SUbstituting `lambda = - (11)/(3)` in equation (1), we get,
`barr*[(2-(11)/(3))hati+(-3+(11)/(3))hatj+4hatk]=1-4((-11)/(3))`
i.e., `barr*[((6-11)/(3))hati+((-9+11)/(3))hatj+4hatk]=(3+44)/(3)`
i.e., `barr*((-5)/(3)hati+(2)/(3)hatj+4hatk)=(47)/(3)`
i.e., `barr*(-5hati + 2hatj + 12hatk) = 47`
This is the required equation of the plane.