Find the equation of the plane passing through the straight line `(x-1)/2=(y+2)/(-3)=z/5` and perpendicular to the plane `x-y+z+2=0.`

Explanation

Let the equation of plane containing the line be
`l(x-1)+m(y+2)+nz=0`. Then
`" "2l-3m+5n=0 and l-m+n=0`
`therefore" "(l)/(2)=(m)/(3)=(n)/(1)`
Hence, the plane is `2(x-1) + 3(y+2)+z =0`
or `" "2x+3y+z+4=0`