QJEEM22Ma5134: JEE Main 2022 Mathematics - Class 12 JEEM22Ma5134 Solution: If two straight lines whose direction cosines are given by the…

Given that the direction cosines satisfy $l + m - n = 0$ , we find that $n = l + m$ .

The other equation is $3l^2 + m^2 + cnl = 0$ , and substituting $n = l + m$ gives $3l^2 + m^2 + cl(l + m) = 0$ .

This simplifies to $(3 + c)l^2 + clm + m^2 = 0$ .

As the lines are parallel, they share the same direction ratios, so we can express $l$ in terms of $m$ , say $l = km$ . Substituting this into our equation gives $(3 + c)(km)^2 + ckm^2 + m^2 = 0$ .

This simplifies to $m^2[k^2(3 + c) + kc + 1] = 0$ .

Since $m \neq 0$ , we must have $k^2(3 + c) + kc + 1 = 0$ . Here, we consider the ratio $k = \frac{l}{m}$ to be constant, since the lines are parallel. The equation then becomes a quadratic equation in $k$ .

As the lines are parallel, the discriminant of the quadratic equation must be equal to zero for the equation to have equal roots. Hence, the discriminant $D = (c^2 - 4(3 + c)) = 0$ .

Solving this quadratic equation gives $c^2 - 4c - 12 = 0$ .

Factoring this equation gives $(c - 6)(c + 2) = 0$ .

Solving for $c$ gives $c = 6, -2$ .

However, we are looking for the positive value of $c$ , so $c = 6$ .

Therefore, the correct answer is 6