1. Interpret the condition

The direction cosines $(l,m,n)$ of the lines satisfy:

$l+m-n=0$ $3l^2+m^2+cnl=0$

Since the question says two straight lines are obtained from these relations and they are parallel, the two possible directions determined by these equations must be the same line of direction. That means the system should give two proportional sets of direction cosines differing only by sign.

2. Use the linear relation

From

$l+m-n=0$

we get

$n=l+m$

Substitute into the quadratic relation:

$3l^2+m^2+cln=0$

so

$3l^2+m^2+cl(l+m)=0$

$3l^2+m^2+cl^2+clm=0$

$(c+3)l^2+clm+m^2=0$

3. Form equation in the ratio $m/l$

Let

$t=\frac{m}{l}$

(If $l=0$, then from $l+m-n=0$, we get $n=m$, and the quadratic gives $m^2=0$, so only trivial solution; hence $l\neq 0$.)

Divide by $l^2$:

$(c+3)+ct+t^2=0$

or

$t^2+ct+(c+3)=0$

This quadratic gives the possible values of $\dfrac{m}{l}$, hence the two possible directions.

4. Condition for parallel lines

For the two lines to be parallel, the two roots must correspond to the same direction ratio (up to overall sign). That happens when the quadratic has equal roots.

So discriminant must be zero:

$c^2-4(c+3)=0$

$c^2-4c-12=0$

Factorize:

$(c-6)(c+2)=0$

Thus,

$c=6 \quad \text{or} \quad c=-2$

The positive value is:

$\boxed{6}$

5. Check with options

Option A: $6$ ✔️

So the correct answer is A.

6. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

They agree.