1. Let $x=f(a),\quad y=f(b),\quad z=f(c),\quad w=f(d).$ Then $x,y,z,w\in\{0,1,2,\dots,10\}$ and since $f$ is one-one, $x,y,z,w$ are all distinct.

2. The given condition is $2f(a)-f(b)+3f(c)+f(d)=0,$ i.e. $2x-y+3z+w=0.$ Rearranging, $y=2x+3z+w.$

3. Since $x,z,w\ge 0$, we have $y=2x+3z+w\ge 0.$ Also $y\le 10$ because codomain is $\{0,1,\dots,10\}$.

   So we must count ordered triples $(x,z,w)$ of distinct elements from $\{0,1,\dots,10\}$ such that $2x+3z+w\le 10,$ and then set $y=2x+3z+w$ with the additional requirement that $y$ is distinct from $x,z,w$.

4. Because of the coefficient $3z$, we must have small values of $z$.

Case 1: $z=0$

Then $y=2x+w$ with distinctness requiring $x,w,0$ distinct, so $x,w\ne 0$ and $x\ne w$. Also $y\le 10$.

Now count ordered pairs $(x,w)$ with $x,w\in\{1,2,\dots,10\}$, $x\ne w$, and $2x+w\le 10.$

For each $x$:

• $x=1$: $w\le 8$, with $w\in\{1,2,\dots,8\}$ and $w\ne 1$ gives $7$ values.
• $x=2$: $w\le 6$, from $\{1,2,\dots,6\}$ excluding $2$ gives $5$ values.
• $x=3$: $w\le 4$, from $\{1,2,3,4\}$ excluding $3$ gives $3$ values.
• $x=4$: $w\le 2$, from $\{1,2\}$ gives $2$ values.
• $x\ge 5$: no solution.

Total so far: $7+5+3+2=17.$

But we must also ensure $y\ne x,w,0$.

• $y=0$ impossible since $2x+w>0$.
• $y=x$ would mean $2x+w=x\Rightarrow x+w=0$, impossible.
• $y=w$ would mean $2x+w=w\Rightarrow x=0$, impossible.

So all $17$ are valid.

Case 2: $z=1$

Then $y=2x+w+3.$ Distinctness requires $x,w\ne 1$, $x\ne w$. Also $2x+w+3\le 10\Rightarrow 2x+w\le 7.$

Now $x$ cannot be $0$? It can be, since only distinct from $z=1$ and $w$. So count ordered pairs $(x,w)$ with $x,w\in\{0,1,\dots,10\}$, $x,w\ne 1$, $x\ne w$, and $2x+w\le 7$.

For each $x$:

• $x=0$: $w\le 7$, $w\in\{0,2,3,4,5,6,7\}$ excluding $w=0$ gives $6$ values.
• $x=2$: $w\le 3$, $w\in\{0,2,3\}$ excluding $2$ gives $2$ values.
• $x=3$: $w\le 1$, possible $w=0$ only, so $1$ value.
• $x\ge 4$: no solution.

So tentative total: $6+2+1=9.$

Now check $y\ne x,w,1$.

• $y=1$ impossible since $2x+w+3\ge 3$.
• $y=x$ gives $2x+w+3=x\Rightarrow x+w+3=0$, impossible.
• $y=w$ gives $2x+w+3=w\Rightarrow 2x+3=0$, impossible.

Hence all $9$ are valid.

Case 3: $z=2$

Then $y=2x+w+6,$ so $2x+w\le 4.$ Also $x,w\ne 2$, $x\ne w$.

Count ordered pairs:

• $x=0$: $w\le 4$, from $\{0,1,3,4\}$ excluding $0$ gives $3$ values.
• $x=1$: $w\le 2$, from $\{0,1\}$ excluding $1$ gives $1$ value.
• $x\ge 2$: no solution.

Tentative total: $3+1=4.$

Check $y\ne x,w,2$.

• $y=2$ impossible.
• $y=x$ impossible.
• $y=w$ would imply $2x+6=0$, impossible.

So all $4$ are valid.

Case 4: $z\ge 3$

Then $3z\ge 9.$ Since $x,w\ge 0$, $y=2x+3z+w\ge 9.$ But if $z=3$, then $2x+w\le 1$. Since $x,w$ must be distinct and neither can equal $3$, only possibility is $(x,w)=(0,1)$ or $(1,0)$. Let us check:

• $(x,w)=(0,1)$ gives $y=10$, valid and distinct.
• $(x,w)=(1,0)$ gives $y=11$, not allowed.

So for $z=3$, there is exactly $1$ solution.

If $z\ge 4$, then $y\ge 12>10,$ so no solution.

Thus Case 4 contributes $1$.

5. Total number of one-one functions is $17+9+4+1=31.$

Therefore, the required number is $\boxed{31}.$