Digits are $1,2,3,4,5,7,9$ Multiple of $11 \rightarrow$ Difference of sum at even \& odd place is divisible by 11 . Let number of the form abcdefg $$ \begin{aligned} & \therefore(a+c+e+g)-(b+d+f)=11 x \\ & a+b+c+d+e+f=31 \\ & \therefore \text { either } a+c+e+g=21 \text { or } 10 \\ & \therefore b+d+f=10 \text { or } 21 \end{aligned} $$ Case-1 $$ \begin{aligned} & a+c+e+g=21 \\ & b+d+f=10 \\ & (b, d, f) \in\{(1,2,7)(2,3,5)(1,4,5)\} \\ & (a, c, e, g) \in\{(1,4,7,9),(3,4,5,9),(2,3,7,9)\} \\ & \therefore \text { Total number in case- } 1=(3!\times 3)(4!)=432 \end{aligned} $$ Case- 2 $$ \begin{aligned} & a+c+e+g=10 \\ & b+d+f=21 \\ & (a, b, e, g) \in\{1,2,3,4)\} \\ & (b, d, f) \&\{(5,7,9)\} \end{aligned} $$ ∴ Total number in case $2=3!\times 4!=144$
Total numbers = 144 + 432 = 576