`P_1 = 2x+3y + z = 1`
`P_2 = x+ 3y + 2z = 2`
Dr of line`_|_`to `P_1 = (2,3,1)`
Dr of line`_|_`to `P_2 = (1,3,2)`
Dr of line L=`(a,b,c)`
`L _|_ P_1, 2a+ 3b + c = 0` (1)
`L _|_ P_2, a+ 3b+2c= 0` (2)
subtracting eqn 2 from 1
`a=c`
`3b= 3a`
so,`a=b=c`
`cos alpha = a/(sqrt(a^2 + b^2+ c^2))= a/sqrt(3a^2) = 1/sqrt3`
option 1 is correct

Correct Answer – d
Since line of intersection is perpendicular to both the planes, direction rations of the line of intersection is ltbgt `|[hati,hatj,hatk],[2,3,1],[1,3,2]|=3hati-3hatj+3hatk`
Hence, `cosalpha=(3)/(sqrt(9+9+9))=(1)/(sqrt3)`