$ (3k,\ 1,\ -5) $$

Since the two lines are perpendicular, the dot product of their direction ratios must be zero.

So,

$$ (-3)(3k) + (2k)(1) + (2)(-5) = 0 $$

$$ -9k + 2k - 10 = 0 $$

$$ -7k - 10 = 0 $$

$$ k = -\frac{10}{7} $$

Now substitute this value of $k$.

Direction ratios of $L_1$ become:

$$ \left(-3,\ -\frac{20}{7},\ 2\right) $$

Multiplying by $7$, we can take the direction vector as:

$$ \vec{a} = (-21,\ -20,\ 14) $$

Direction ratios of $L_2$ become:

$$ \left(-\frac{30}{7},\ 1,\ -5\right) $$

Multiplying by $7$, we can take the direction vector as:

$$ \vec{b} = (-30,\ 7,\ -35) $$

The required plane contains $L_1$ and is parallel to $L_2$.

So the plane is parallel to both vectors $\vec{a}$ and $\vec{b}$.

Hence, its normal vector is:

$$ \vec{n} = \vec{a} \times \vec{b} $$

$$ \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -21 & -20 & 14 \\ -30 & 7 & -35 \end{vmatrix} $$

$$ \vec{n} = (602,\ -1155,\ -747) $$

The plane contains the line $L_1$, so it passes through the point $(1,2,3)$.

Using the plane equation,

$$ 602(x-1) -1155(y-2) -747(z-3) = 0 $$

Expanding,

$$ 602x - 602 -1155y + 2310 -747z + 2241 = 0 $$

$$ 602x -1155y -747z + 3949 = 0 $$

So the correct option is:

$$ \boxed{\text{Option B } \; 602x - 1155y - 747z + 3949 = 0} $$

Three Dimensional Geometry

Continue this paper