Find the equation of the plane passing through the point (-1, 3, 2) and perpendicular to each of the planes `x+2y+3z=5\ ` and `3x+3y+z=0`

Explanation

Equation of plane passing through the point (-1,3,2) is
`a[x-(-1)]+b(y-3)+c(z-2)=0`
or `a(x+1)+b(y-3)+c(z-2)=0"….."(1)`
Plane (1) is perpendicular to the plane `x+2y+3z=5`.
`:. a*1+b*2+c*3 = 0 rArr 3a+3b+c=0"….."(3)`
Solving equations (2) and (3) by cross multiplication method.
`(a)/(2xx1-3xx3) = (b)/(3xx3-1xx1)=(c )/(1xx3-2xx3)`
`rArr (a)/(-7) = (b)/(8) = (c )/(-3) = lambda`
`rArr a = - 7lambda, b = 8lambda, c = - 3lambda`
Put the values of `a,b` and c in equation(1),
`-7lambda (x+1)+8lambda(y-3) 3lambda(z" "2) = 0`
`rArr (-7x-7) + (8y-24) - 3z + 6 = 0`
`rArr -7x+8y-3z-25 = 0`
`rArr 7x-8y+3z+25=0`
Therefore, equation of required plane is `7x-8y+3z+25=0`.