Let L be the line belonging to the family of straight lines $(a+2b)x+(a-3b)y+a-8b=0,a,b\in R,$ which is the farthest from the point (2,2).If L is concurrent with the lines $x-2y+1=0$ and $3x-4y+\lambda =0,$ then the value of $\lambda$ is

Given lines $(x+y+1)+b(2x-3y-8)=0$ are concurrent at the point of intersection of the lines $x+y+1=0$ and $2x-3y-8=0,$ which is (1,-2). Now, the line through A(1,-2), which is farthest from the point B(2,2), is perpendicular to AB. Now, the slope of AB is 4. Then the required line is $y+2=-(1/4)(x-1)orx+4y+7=0.$ Lines $x+4y+7=0$ and $x-2y+1=0$ intersect at (-3,-1) which must satisfy the line $3x-4y+\lambda =0.$ Then $-9+4+\lambda =0or\lambda =5$