The equation of the plane containing the line $\frac{x-\alpha}{I}=\frac{y-\beta}{m}=\frac{z-\gamma}{n}$ is $a(x-\alpha)+b(y-\beta)+c(z-\gamma)=0$, where $a l+b m+c n$ is equal to

0

Since straight line lies in the plane so it will be perpendicular to the normal at the given plane. Since direction cosines of straight line are l, m, n and direction ratios of normal to the plane are a, b, c. So, al + bm + cn = 0.

Hence (4) is the correct answer.