If three distinct squares $\left(1\times 1\right)$ size are chosen at random from a chess board, then the probability that they all do not  lie on a diagonal line is

fThere are $64$ squares on a chessboard.

Total number of ways of selecting three out these squares ${=}^{64}{C}_3$.

We need to find the probability of selecting three squares such that all three of them do not lie on a diagonal of the chessboard together.

$\Rightarrow P(All not on diagonal) = 1 - P(All are on diagonal)$

Now, the diagonals of a chessboard are as shown in the figure:

Out of these $15$ diagonals, the two at top-left and the two at bottom-right are not to be considered as the number of squares lying along them are less than three.

For diagonals above the middle-longest diagonal, number of cases where three squares are selected, all on a diagonal,

${=}^7{C}_3{+}^6{C}_3{+}^5{C}_3{+}^4{C}_3{+}^3{C}_3$ $= 70$

Now, the diagonals below the middle diagonal are identical to the ones above it. 

Hence total required cases below the middle diagonal are equal to the cases above it.

For the middle diagonal, number of cases ${=}^8{C}_3=56$.

Thus, total number of cases for this set of diagonals $= 2\times 70 + 56 = 196$.

Now, symmetrical to this set of diagonals, there is another set perpendicular to this one from which squares can be chosen.

Hence, finally, total number of cases for choosing three squares, all on a diagonal $= 2\times 196 = 392$.

Thus,

 $$\begin{gathered} P(All not on diagonal) = 1 - \frac{Total cases where all are on diagonal}{Total possible cases} \\ =1- \frac{392}{{ }^{64}{C}_3} \\ = 1 - \frac{7}{744} \\ =\frac{737}{744} \end{gathered}$$

Hence, option $2$ is correct.