Correct Answer - D
`a=sqrt(6)-sqrt(5)`
`b=sqrt(5)-2`
`c=2-sqrt(3)`
Rationalize all the terms
`a=(sqrt(6)-sqrt(5))/(sqrt(6)+sqrt(5))xxsqrt(6)+sqrt(5)`
`=1/(sqrt(6)+sqrt(5))`
`b=(sqrt(5)-2)/(sqrt(5)+2)xxsqrt(5)+2=1/(sqrt(5)+2)`
`=1/(sqrt(5)+sqrt(4))`
`c=(2-sqrt(3))/(2+sqrt(3))xx 2+sqrt(3)`
`=1/(2+sqrt(3))=1/(sqrt(4)+sqrt(3))`
NOw, `a=1/(sqrt(6)+sqrt(5))`
`b=1/(sqrt(5)+sqrt(4))`
`c=1/(sqrt(4)+sqrt(3))`
Now the term whose denominator is largest is the smallest term.
So `altbltc`.