question about division

If $N|Ax+By$ implies $N|Cx+Dy$ then also $N|CAx+CBy$ and $N|CAx + DAy$, hence $N|(AD-BC)y$ and so $N|AD-BC$. For example, in the first question answer $C$, $7 \cdot 4 + 3 \cdot 9 = 55$ is divisible by $11$.

Another way of looking at that is that $(C,D)$ must be some multiple of $(A,B)$. So $C/A = D/B$, which can be tested via $CB=AD$.

If you have more than two variables, say $N|Ax+By+Cz$ implies $N|Dx+Ey+Fz$ then the same method gives $A/D=B/E=C/F$ and so $AE=BD$ and $BF=CE$. Try it on the second question.

Hint $\ \mathbb F = \mathbb Z/11\:$ & $\:\mathbb Z/13\,$ are fields, so $\rm\,\mathbb F^{\:\!n}$ is a vector space, hence lines through the origin are uniquely determined by any non-origin point on the line. To find such a point, pick an easily invertible coefficient $c$ of the equation and scale the equation by $c^{-1}$ so the coefficient becomes $1$.

E.g. $\rm\ \color{#c00}4x -5y + 2z = 0\:$ over $\rm\:\mathbb Z/13.\:$ We look for an easy invertible coef, e.g. one dividing $13\pm1,\,$ e.g. $\,\color{#c00}4\cdot 3 = 13- 1 = -1,\,$ so scaling by $\,4^{-1} = - 3$ yields $\rm\: x +2y - 6z = 0,\,$ so $\rm\,x = -2y+6z.\,$ Picking "simple" values for $\rm\:y,z\:,\:$ say $\rm\:y=0,\:z=2\:$ yields $\rm\:x=-1,\:$ which lies only on the line $\rm\:(D)\ -7x +12y+3z = 0.$

This method is both general and quick - esp. quick when there is an easily invertible coefficient dividing modulus $\pm 1\:$ (which occurs frequently for small moduli by the law of small numbers).

There are general methods, and in this case there is even a quick one. But in general the "specific numbers" strategy you mentioned is well-suited for this kind of test, in which you have to solve too many too easy problems in too little time.

Below is my strategy. Real multiple choice test survivors can undoubtedly supply better ones.

For the first problem, I would find some particular numbers $x$ and $y$, not both $0$, such that $3x+7y \equiv 0 \pmod{11}$. My immediate choice is the no thinking $x=7$, $y=-3$. This quickly kills all possibilities except D. (I might kill B and E by using $x=y=0$.)

For the second problem, same thing. Let $x=4$, $y=5$, $z=0$. The $z=0$ is to forget about $z$. Got unlucky, D and E are still alive. So pick $x=0$, $y=2$, $z=5$. That kills E.

As to general methods, for linear algebra reasons, if we know that $ax+by\equiv 0\pmod{p}$, where $p$ is prime, we cannot have $cx+dy \equiv 0 \pmod{p}$, with the unless $ad-cb \equiv 0\pmod{p}$. With some care, the idea can be extended to non-prime moduli. So a few quick determinant calculations modulo $11$ also give the answer to the first problem. (The constant terms in B and E kill these as choices, so we needn't bother with them.)

In the second problem, since any variable could be set to $0$, there cannot be a solution unless all $2$ variable problems obtained by setting one variable to $0$ have a solution. This brings us back to calculations like those in the preceding paragraph. But to rule out some possibilities, we may have to do more than one determinant calculation.

My preference is substitution rather than determinant calculations. From the point of view of arithmetic, they are of equal difficulty, indeed identical. But substitution is something you have better intuition about, it is "theory-free."

Added: In the $3x+7y$ problem, I immediately jumped to using $x=7$, $y=-3$. We could replace $7$ by $-4$, and use $x=3$, $y=4$. We could in addition replace $3$ by $-8$, obtaining $(-8x)+(-4y)$, which is equivalent to $2x+y$, and we can use $x=1$, $y=-2$. These changes make the subsequent arithmetic simpler. Should we "invest" in such simplifications? It depends on how comfortable we are with mental arithmetic. My gut reaction is not to invest, I could make a minus sign error, and blow everything. Act, don't think, and go on to the next question.