The number of 7 digit numbers which are multiples of 11 and are formed using all the digits 1, 2, 3, 4, 6, 7, 8 is 

Now,  $(a+c+e+g)-(b+d+f)=11x$
And  $a+b+c+d+e+f+g=31$
Case (i) 
$a+c+e+g=21$ and  $b+d+f=10$
$b+d+f=10 \Rightarrow (b,d,f)=(1,2,7)(1,3,6)$  
Total number of cases $=2(3!)(4!)\Rightarrow 228$
Case (ii)
$a+c+e+g=10$ and  $b+d+f=21$
$(b,d,f)=(6,7,8)$ possible number of cases  $=(3!)(4!)=144$
Finally, total number of cases $= 432$