The equation of a line on the plane $ x + y + z + = 1$ such that line $\frac{x-1}{2} = \frac{y-1}{1} = \frac{z-1}{1} $ and the required line form a plane which is perpendicular to the plane $ x + y + z + = 1$ is :
My Approach : Let any point on the given line $\frac{x-1}{2} = \frac{y-1}{1} = \frac{z-1}{1} =r$ is $(2r+1, r+1, r+1)$ ...... So by this $r=(-1/2)$. And another point is $(1,1,1)$ .
After this I am not able to come to the final answer .... Pleas explain and tell the correct approach.
$(2,1,1)$ is a direction vector of the given line.
Let $(p,q,r)$ be a direction vector of the required line.
Then $(p,q,r)\cdot (2,1,1)=0$.
$(p,q,r)\times (2,1,1)$ is a normal vector of the plane containing the two lines. As this plane is perpendicular to the plane $x+y+z=1$,
$$(1,1,1)\cdot[(p,q,r)\times (2,1,1)]=0$$
So we can find $p:q:r$.
Put $(2t+1,t+1,t+1)$ into $(x,y,z)$ to find the point of intersection of the given line and the given plane, the point is also on the required line.
We are now ready to give the equation of the line.