Equation of the plane that contains the line $\frac{x-1}{1}=\frac{y+1}{2}=\frac{z-1}{1}$ and perpendicular to the plane 3x - y - z + 7 = 0 is:

x - 4y + 7z - 12 = 0

A Plane containing the line:

(x-1)/1 = (y+1)/(2) = (z-1)/1 is

a(x - 1) + b(y + 1 ) + c(z - 1 ) = 0 …. … .. (1) , satisfying the condition ;

 a  +2b + c = 0 …. ….. ….(2) .

Since the plane (1) is perpendicular to another given plane 3x - y - z + 7 = 0 , therefore the normals to these two planes must also be perpendicular to each other, therefore we must have ;

3a  - b - c = 0 …. ….. …. …. (3) .

Solving eqns. (2) & (3) by cross multiplication method, we get a = - 1 , b = 4 & c = -7 . Putting these values of a, b , c in eq. (1) , we get the required equation of the plane as ;

x- 4y +7z + -12 = 0 .