Happiness is Mandatory

Question

If you're sad, that's wrong. We need to scrub out ALL sadness.

This is a sad face, for reference:

  ......  
  .S..S.  
 ........ 
 ...SS... 
..SS..SS..
.S......S.
 ........ 

Where S is a consistent pixel, . is any pixel other than S, and is any pixel. Note that the sad face is not a perfect square, only the . should be considered. Theoretical sad face overlap in the corners is possible, but overwriting one sad face has a chance to eliminate the other, e.g. if it was in a . character and is overwritten with an S.

Input

An image, in any acceptable format. I would recommend a color matrix.

Output

The same image, but take all those treasonous sad faces and replace their S characters with randomized data. The data should be either random RGB values, or values taken from the palette of the image. You just need to have a nonzero probability of all outputs; you do not need to make it uniform.

Rules

• S can be any pixel, as long as it's the same pixel throughout the formation.
• Remove ALL sad faces. If the removal makes a new sad face, then remove that too.
• You should only modify S pixels. Unauthorized modification of pixels is insubordination, punishable by disqualification.
• The shortest answer in bytes wins.

Test Cases

These have one sad face each:

• 
• 

This has 2 sad faces, on the left and right:

This has 2 sad faces overlapping:

This has 2 sad faces, in the top-left and bottom-right; the bottom-left and top-right have S characters in the place of . characters:

Answer 1

C (gcc), 339 337 331 323 bytes

• Saved two eight fifteen sixteen bytes thanks to ceilingcat.

C,a,l,A,m,i,t,o,u,s;n(e,z,Z)int*e;{for(C=1;C;C+=i)for(C=s=0;s<Z;s++)for(u=0;u<z;u++)if(u>8&s>5)for(m=e[~-u+~-s*z],A=i=1;++A<4;)for(o=0;o<7;o++)for(t=0;t<10;t++)a=e[l=u+t-9+(s+o-6)*z],(o-1||t-3&&t-6)*(o-3||t<4|5<t)*(o-4||t<2|3<t&t<6|7<t)*(o-5||t-1&&t-8)?o>1|t>1&t<8&o<2|o>3&o<6|t&&t<9&&(i*=a-m):A<3?i*=a==m:i?e[l]=rand():e;}

Try it online!