Find the equations of the line passing through the point `(3,0,1)` parallel to the planes `x+2y=0` and `3y-z=0`.

Explanation

Let the direction rations of the required line be a,b,c
Then its equation is given by
`(x-3)/(a)=(y-0)/(b) =(z-1)/(c )`
Since the line `(i)` is parallel to each of the planes `xd+2y+oz=0` and `0x+3y-z=0` so it must be perpendicular to the normal of each of these planes
`therefore axx1+bxx2+cxx0=0 rarr a+2b+0c =0` and `axx0 +bxx3+cxx(-1)=0 rarrr 0a + 3b -c =0`
on solving `(ii)` and `(iii)` by cross multiplication we get
`(a)/(-2-0)=(b)/(0+1)=(c )/(3-0)`
`rarr (a)/(-2)=(b)/(1)=(c )/(3) =lambda(say) rarr a= 3 lambda , b = lambda` and `c= 3lambda`
putting these values of `a,b,c` in `(i)` we get
`(x-3)/(-2lambda)=(y-0)/(lambda)=(z-1)/(3 lambda) rarr (x-3)/(-2)=(y)/(1)=(z-1)/(3)`
Hence the required equation of the line is `(x-3)/(-2)=(y)/(1)=(z-1)/(3)`