JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Probability

A six-digit number uses six of the seven digits, so exactly one digit is left out. The full digit sum is $21$, so the six chosen digits add to a multiple of $3$ precisely when the omitted digit is $0$, $3$ or $6$.
Omitting $0$ leaves $\{1,2,3,4,5,6\}$ and all $6!=720$ arrangements are six-digit numbers. Omitting $3$, or omitting $6$, leaves a set containing $0$, which cannot occupy the first place, so each contributes $6!-5!=600$. The favourable count is $720+600+600=1920$.
Counting all six-digit numbers the same way, omitting $0$ gives $720$ and each of the other six choices gives $600$, so the total is $720+6(600)=4320$. Hence

The probability of the event is $\frac{4}{9}$.