The given inequality $3(x-1) \leq 2(x-3)$
=> $ 3(x-1) \leq 2x-6$
adding 3 on both sides of inequality
=> $ 3x-3+3 \leq 2x-6+3$
=> $3x \leq 2x-3$
Subtracting 2x on both sides,
$3x-2x \leq 2x-6+3$
=> $3x \leq 2x \leq 2x -3 -2x$
$=> x \leq -3$
All real numbers which are less then and equal to -3 satisfy the given in equality.
The solution set is $(- \infty, -3]$
Hence B is the correct answer.