1. Given data

We have:

• Point on incident ray: $P(3,10)$
• Mirror line: $2x+y=6$
• Reflected ray passes through $Q(7,2)$

We need the incident ray equation in the form $ax+by+1=0$ and then compute $a^2+b^2+3ab.$

2. Use the reflection property

A standard method is to reflect the point $Q(7,2)$ across the mirror line $2x+y-6=0$. If $Q'$ is the reflection of $Q$, then the incident ray is simply the straight line joining $P$ and $Q'$.

3. Find reflection of $Q(7,2)$ in the line $2x+y-6=0$

For reflection of point $(x_0,y_0)$ across line $Ax+By+C=0$: $x' = x_0 - \frac{2A(Ax_0+By_0+C)}{A^2+B^2},$ $y' = y_0 - \frac{2B(Ax_0+By_0+C)}{A^2+B^2}.$

Here, $A=2,\quad B=1,\quad C=-6,$ $(x_0,y_0)=(7,2).$

First compute: $Ax_0+By_0+C = 2\cdot 7 + 1\cdot 2 - 6 = 14+2-6=10.$ Also, $A^2+B^2=4+1=5.$

So, $x' = 7 - \frac{2\cdot 2\cdot 10}{5}=7-8=-1,$ $y' = 2 - \frac{2\cdot 1\cdot 10}{5}=2-4=-2.$

Thus the reflected point is $Q'(-1,-2).$

4. Equation of incident ray

The incident ray passes through $P(3,10)$ and $Q'(-1,-2)$.

Slope of line $PQ'$ is $m=\frac{10-(-2)}{3-(-1)}=\frac{12}{4}=3.$

Hence equation is $y-10=3(x-3).$

Simplifying: $y-10=3x-9$ $y=3x+1$ $3x-y+1=0.$

Therefore, $a=3,\quad b=-1.$

5. Compute the required expression

$a^2+b^2+3ab = 3^2+(-1)^2+3(3)(-1).$

$=9+1-9=1.$

6. Comparison with stored answer

Our derived answer is $\boxed{1}$ which matches the stored correct answer.