1. Key Concepts and Formulas

• Coplanarity of Two Lines: Two lines, $L_1: \vec r = \vec a_1 + t\vec b_1$ and $L_2: \vec r = \vec a_2 + s\vec b_2$, are coplanar if the scalar triple product of the vector connecting a point on $L_1$ to a point on $L_2$, and their respective direction vectors, is zero. This condition is expressed as: $(\vec a_2 - \vec a_1) \cdot (\vec b_1 \times \vec b_2) = 0$ In Cartesian form, if $A_1(x_1, y_1, z_1)$ and $A_2(x_2, y_2, z_2)$ are points on $L_1$ and $L_2$ respectively, and $\vec b_1 = (l_1, m_1, n_1)$ and $\vec b_2 = (l_2, m_2, n_2)$ are their direction vectors, the condition is: $\begin{vmatrix} x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \end{vmatrix} = 0$
• Equation of a Plane Containing Two Coplanar Lines: If two lines are coplanar and non-parallel (meaning they intersect), the equation of the plane containing them can be found using a point from one of the lines and the normal vector to the plane. The normal vector $\vec{n}$ is perpendicular to both direction vectors, so $\vec{n}$ is parallel to $\vec b_1 \times \vec b_2$. The equation of a plane passing through a point $(x_0, y_0, z_0)$ with normal vector $(A, B, C)$ is $A(x-x_0) + B(y-y_0) + C(z-z_0) = 0$. Alternatively, if the lines are coplanar, the plane equation can be directly written as: $\begin{vmatrix} x - x_1 & y - y_1 & z - z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \end{vmatrix} = 0$
• Checking if a Point Lies on a Plane: A point $(x_p, y_p, z_p)$ lies on a plane $Ax + By + Cz + D = 0$ if, upon substitution, it satisfies the equation, i.e., $Ax_p + By_p + Cz_p + D = 0$.

2. Step-by-Step Solution

Step 1: Extract Information from the Given Line Equations

The given lines are in symmetric form: $L_1: \frac{x-1}{\lambda}=\frac{y-2}{1}=\frac{z-3}{2}$ $L_2: \frac{x+26}{-2}=\frac{y+18}{3}=\frac{z+28}{\lambda}$

From $L_1$:

• A point on $L_1$ is $A_1 = (1, 2, 3)$.
• The direction vector of $L_1$ is $\vec b_1 = (\lambda, 1, 2)$.

From $L_2$:

• A point on $L_2$ is $A_2 = (-26, -18, -28)$.
• The direction vector of $L_2$ is $\vec b_2 = (-2, 3, \lambda)$.

Now, we find the vector connecting $A_1$ and $A_2$: $\vec{A_1A_2} = A_2 - A_1 = (-26 - 1, -18 - 2, -28 - 3) = (-27, -20, -31)$.

Step 2: Apply the Coplanarity Condition to find $\lambda$

Since the lines are coplanar, we use the determinant condition: $\begin{vmatrix} x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \end{vmatrix} = 0$ Substituting the extracted values: $\begin{vmatrix} -27 & -20 & -31 \\ \lambda & 1 & 2 \\ -2 & 3 & \lambda \end{vmatrix} = 0$ Expand the determinant: $-27(1 \cdot \lambda - 2 \cdot 3) - (-20)(\lambda \cdot \lambda - 2 \cdot (-2)) + (-31)(\lambda \cdot 3 - 1 \cdot (-2)) = 0$ $-27(\lambda - 6) + 20(\lambda^2 + 4) - 31(3\lambda + 2) = 0$ $-27\lambda + 162 + 20\lambda^2 + 80 - 93\lambda - 62 = 0$ Combine like terms: $20\lambda^2 + (-27 - 93)\lambda + (162 + 80 - 62) = 0$ $20\lambda^2 - 120\lambda + 180 = 0$ Divide the entire equation by 20: $\lambda^2 - 6\lambda + 9 = 0$ This is a perfect square trinomial: $(\lambda - 3)^2 = 0$ Therefore, $\lambda = 3$.

Step 3: Find the Equation of the Plane P

With $\lambda = 3$, the direction vectors of the lines are: $\vec b_1 = (3, 1, 2)$ $\vec b_2 = (-2, 3, 3)$

The normal vector $\vec n$ to the plane P is parallel to $\vec b_1 \times \vec b_2$: $\vec n = \vec b_1 \times \vec b_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 1 & 2 \\ -2 & 3 & 3 \end{vmatrix}$ $\vec n = \mathbf{i}(1 \cdot 3 - 2 \cdot 3) - \mathbf{j}(3 \cdot 3 - 2 \cdot (-2)) + \mathbf{k}(3 \cdot 3 - 1 \cdot (-2))$ $\vec n = \mathbf{i}(3 - 6) - \mathbf{j}(9 + 4) + \mathbf{k}(9 + 2)$ $\vec n = -3\mathbf{i} - 13\mathbf{j} + 11\mathbf{k}$ So, the direction ratios of the normal vector are $(-3, -13, 11)$.

We can use the point $A_1(1, 2, 3)$ and the normal vector $(-3, -13, 11)$ to write the equation of plane P: $-3(x - 1) - 13(y - 2) + 11(z - 3) = 0$ $-3x + 3 - 13y + 26 + 11z - 33 = 0$ $-3x - 13y + 11z + (3 + 26 - 33) = 0$ $-3x - 13y + 11z - 4 = 0$ Multiplying by -1 to get positive leading coefficients: $3x + 13y - 11z + 4 = 0$ This is the equation of plane P.

Step 4: Check Which Point Does NOT Lie on Plane P

We substitute the coordinates of each given option into the plane equation $3x + 13y - 11z + 4 = 0$.

• (A) $(0, -2, -2)$: $3(0) + 13(-2) - 11(-2) + 4 = 0 - 26 + 22 + 4 = 0$. This point lies on the plane.

• (B) $(-5, 0, -1)$: $3(-5) + 13(0) - 11(-1) + 4 = -15 + 0 + 11 + 4 = 0$. This point lies on the plane.

• (C) $(3, -1, 0)$: $3(3) + 13(-1) - 11(0) + 4 = 9 - 13 + 0 + 4 = 0$. This point lies on the plane.

• (D) $(0, 4, 5)$: $3(0) + 13(4) - 11(5) + 4 = 0 + 52 - 55 + 4 = 1 \neq 0$. This point does NOT lie on the plane.

Therefore, the point $(0, 4, 5)$ does not lie on plane P.

3. Common Mistakes & Tips

• Sign Errors: Be extremely careful when extracting points from line equations (e.g., $x+26$ means $x-(-26)$) and when expanding determinants or cross products. A single sign error can lead to an incorrect value of $\lambda$ or an incorrect plane equation.
• Determinant Expansion: Double-check the expansion of the $3 \times 3$ determinant. It's a common source of calculation errors.
• Checking Coplanarity for Parallel Lines: If direction vectors were proportional, the lines would be parallel. In that case, the cross product $\vec b_1 \times \vec b_2$ would be $\vec 0$. The coplanarity condition would then become $(\vec a_2 - \vec a_1) \cdot \vec b_1 = 0$ (if lines are parallel and coplanar, the vector connecting points on them must be perpendicular to the common direction vector). Our lines were not parallel, so the general condition was appropriate.

4. Summary

We first used the coplanarity condition for two lines to determine the unknown parameter $\lambda$. This involved setting the scalar triple product of the vector connecting points on the lines and their direction vectors to zero, which simplified to a quadratic equation for $\lambda$. Solving this equation yielded $\lambda=3$. Next, we found the equation of the plane P containing these two lines by using a point from one of the lines and the normal vector, which was obtained by taking the cross product of the two direction vectors. Finally, we checked each given option by substituting its coordinates into the plane equation. The point that did not satisfy the equation was the answer. Our calculations showed that the plane is $3x + 13y - 11z + 4 = 0$, and the point $(0, 4, 5)$ does not lie on this plane.

The final answer is $\boxed{(0,4,5)}$ which corresponds to option (D).