If the distance between the plane Ax 2y + z = d and the plane containing the lines 2 1x = 3 2y = 4 3z and 3 2x = 4 3y = 5 4z is 6 , then |d| is

Answer

`6`

Explanation

Let normal to plane be `lhati+mhatj+nhatk`
`" "2l+3m+4n=0`
and `" "3l+4m+5n =0`
`" "(l)/(-1) = (m)/(2)= (n)/(-1)`
Equation of plane will be
`" "a(x-1)+b (y-2)+c(z-3)=0`
or `" "-1(x-1)+2(y-2)-1(z-3)=0`
or `" "-x+1+2y-4-z+3=0`
or `" "-x+2y+z=0`
or `" "x-2y+z=0`
or `" "(|d|)/(sqrt6) = sqrt6`
or `" "d=6`