Number of roots

$\begin{aligned} a^2+b^2+c^2+d^2+12 & < ab + 3b+6c+2d \\ a^2-ab+b^2-3b+c^2-6c+d^2-2d+12 & < 0 \\ a^2-ab+\frac {b^2}4+\frac 34(b^2-4b+4)+c^2-6c+9+d^2-2d+1-1 & < 0 \\ \left(a-\frac b2\right)^2+\frac 34(b-2)^2+(c-3)^2+(d-1)^2 & < 1 \end{aligned}$

We note that the LHS $\le 0$ and equality occurs when $a=1$ , $b=2$ , $c=3$ and $d=1$ . Since the LHS is integral, other sets of $(a,b,c,d)$ will make it $\ge 1$ . Therefore, there is only $\boxed{1}$ solution.

Since $a,b,c,d$ are all positive integers, we will find $(a,b,c,d)$ such that

${ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+{ d }^{ 2 }+13\le ab+3b+6c+2d\\ \Longleftrightarrow { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+{ d }^{ 2 }+13-ab+3b+6c+2d\le 0\\ \Longleftrightarrow { \left( a-\frac { 1 }{ 2 } b \right) }^{ 2 }+3{ \left( \frac { 1 }{ 2 } b-1 \right) }^{ 2 }+{ \left( c-3 \right) }^{ 2 }+{ \left( d-1 \right) }^{ 2 }\le 0\\ \Longleftrightarrow a=1,b=2,c=3,d=1.$

Hence, there is $\boxed { 1 }$ set $(a,b,c,d)$ .