Off by One

$a = \dfrac{X-1}{2} \quad b = \dfrac{X-2}{3} \quad c =\dfrac{X-3}{4}$ $a,b,c$ are positive integers if $X$ is the odd integer less by 1 than that of $\text{l.c.m} (2,3,4)$ . Since $\text{l.c.m}(2,3,4)$ is $12$ and $X = 12-1 =11$ .

$a$ , $b$ and $c$ has positive integer solution if $k\cdot\,( \text{l.c.m}(2,3,4)) -1 , k\in\mathbb N$ .

$1. X = 2a+1$

$2. X= 3b+2$

$3. X=4c+3$

Actually $3.$ it is telling the same and more information that $1.$ so we can ignore it. Equaling $2$ and $3$ we have a super simple Diophantine Equation that has solutions $b=3$ and $c=2$ , substituting, $X=11$

$X\equiv-1(mod2),X\equiv-1(mod3),X\equiv-1(mod4)$ .So $X\equiv-1(mod12)$ ,smallest positive X is 12-1=11

4c+3 is going to be the largest value for the smallest integer so we can use trial and error to check. If c=1, x=7 which does not give an integer value for b. When c=2, all values are integers and X=11.