As Michael E2's comment already says: How you grade depends on what you want to achieve. One goal might be, that pure guessing should have the same mean value as not doing anything at all.
In a yes-no-scheme like in Dirk's answer, that is easy. Every wrong answer gives the same points negativ, which the correct answer gives positive.
If only correct answers are to be marked (and wrong answers are indistinguishable from no answer), this needs to be different.
Suppose a question has $t$ correct and $f$ wrong answer options and the question has a weight of $w$, which means giving exactly the correct answers yields $w$ points. Then each correct mark gives $\frac{w}{t}$ points and each wrong mark gives $-\frac{w}{f}$ points.
Example: I take your example. The weight is $1$, there are $t=3$ correct answers and $f=7$ wrong answers. So each correct mark gives $\frac{1}{3}$ point, each wrong mark gives $-\frac{1}{7}$ point. The students marks all three correct options and one wrong option, so he should get $3\cdot\frac{1}{3}+1\cdot\left(-\frac{1}{7}\right)=\frac{6}{7}$ points.
Reasoning: If you don't discriminate the answer options, then you get the following crosstab for the points to be awarded:
| |mark|no mark|
----------------------
|correct| x | 0 |
|wrong | y | 0 |
The 0
on the right side are necessary, because you don't award someone points (neither positive nor negative) for not answering. The $y$ is gonna be negative, of course.
Let's take a random guesser. Each correct options gives him an expected value of $\frac{x}{2}$. Each wrong option gives him an expected value of $\frac{y}{2}$. Suppose, that the random guesser answers each option independantly (by no way realistic if $t=1$ is not known). Than the expected value of the whole question is $E=t\cdot\frac{x}{2}+f\cdot\frac{y}{2}$. This should be $0$, so you get
$$tx+fy=0$$
Furthermore, giving exactly the correct answers earns $tx$ points, which should be the weight $w$:
$$tx=w$$
Solving these equations for $x$ and $y$ yields:
$$x=\frac{w}{t},\quad y=-\frac{w}{f}=-\frac{t}{f}x$$