Key Concepts and Formulas

• Equation of a Line: A line passing through a point $P_0(x_0, y_0, z_0)$ and having a direction vector $\vec{d} = \langle a, b, c \rangle$ can be written in symmetric form as $\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}$.
• Condition for a Point on a Line: A point $P_1(x_1, y_1, z_1)$ lies on a given line if its coordinates satisfy the line's equation. For a symmetric equation, substituting the coordinates of $P_1$ must result in equal ratios.
• Condition for a Plane Containing a Line: A plane with equation $Ax+By+Cz=D$ (and normal vector $\vec{n} = \langle A, B, C \rangle$) contains a line (with direction vector $\vec{d}$) if and only if:
  1. Any point on the line (e.g., $P_0$) satisfies the plane's equation.
  2. The normal vector of the plane is perpendicular to the direction vector of the line, i.e., their dot product is zero: $\vec{n} \cdot \vec{d} = 0$.
• Equation of a Plane through a Point and a Line (General Case): If a given point $P_1$ does NOT lie on the line $L$, then the plane is uniquely defined. Let $P_0$ be a point on $L$ and $\vec{d}$ be the direction vector of $L$. The normal vector to the plane will be $\vec{n} = \vec{P_0 P_1} \times \vec{d}$. The equation of the plane can then be found using $\vec{n} \cdot (\vec{r} - \vec{P_1}) = 0$ or $\vec{n} \cdot (\vec{r} - \vec{P_0}) = 0$.

Step-by-Step Solution

Step 1: Identify the given information. We are given a point and a line.

• The given point is $P_1 = (3,2,0)$.
• The given line is $L: \frac{x - 4}{1} = \frac{y - 7}{5} = \frac{z - 4}{4}$.
  • From the symmetric form of the line, we can identify a point on the line, $P_0 = (4,7,4)$.
  • We can also identify the direction vector of the line, $\vec{d} = \langle 1, 5, 4 \rangle$.

Step 2: Verify if the given point $P_1$ lies on the line $L$.

• Why this step? This initial check is crucial. If the point lies on the line, the problem simplifies significantly, as we are then just looking for any plane that contains the given line. If the point does not lie on the line, we would use the cross product method to find the plane's normal vector.
• Substitute the coordinates of $P_1(3,2,0)$ into the symmetric equation of the line:
  • For the x-coordinate: $\frac{3 - 4}{1} = \frac{-1}{1} = -1$.
  • For the y-coordinate: $\frac{2 - 7}{5} = \frac{-5}{5} = -1$.
  • For the z-coordinate: $\frac{0 - 4}{4} = \frac{-4}{4} = -1$.
• Since all three ratios are equal to $-1$, the point $P_1(3,2,0)$ does lie on the given line $L$.

Step 3: Determine the implication and choose an efficient strategy.

• Implication: Because $P_1(3,2,0)$ lies on the line $L$, we are simply looking for a plane that contains the entire line $L$. Any such plane will automatically contain $P_1$. This means the problem reduces to finding which option represents a plane that contains the given line.
• Strategy: To check which of the given options contains the line $L$, we can use the property that if a plane contains a line, it must contain every point on that line. We already have two distinct points on the line: $P_1(3,2,0)$ and $P_0(4,7,4)$. We will test these points against the equations of the planes in the options. Alternatively, we can check if the plane's normal vector is perpendicular to the line's direction vector, and if one point from the line lies in the plane. For multiple-choice questions, point substitution is often the fastest.

Step 4: Test the given options.

• Option (A): $x-y+z=1$

  • Check with $P_1(3,2,0)$: $3 - 2 + 0 = 1$. This is true.
  • Check with $P_0(4,7,4)$: $4 - 7 + 4 = 1$. This is true.
  • Since the plane $x-y+z=1$ contains two distinct points $P_1$ and $P_0$ from the line $L$, it must contain the entire line $L$. This is a plausible answer.
  • (Optional Verification using normal vector): The normal vector for plane (A) is $\vec{n_A} = \langle 1, -1, 1 \rangle$. The direction vector of the line is $\vec{d} = \langle 1, 5, 4 \rangle$.
    • Calculate the dot product: $\vec{n_A} \cdot \vec{d} = (1)(1) + (-1)(5) + (1)(4) = 1 - 5 + 4 = 0$.
    • Since the dot product is zero, $\vec{n_A}$ is perpendicular to $\vec{d}$. This, combined with $P_1$ lying in the plane, confirms that the line $L$ lies entirely in plane (A).

• Option (B): $x+y+z=5$

  • Check with $P_1(3,2,0)$: $3 + 2 + 0 = 5$. This is true.
  • Check with $P_0(4,7,4)$: $4 + 7 + 4 = 15 \neq 5$. This is false.
  • Therefore, plane (B) does not contain the line $L$.

• Option (C): $x+2y-z=1$

  • Check with $P_1(3,2,0)$: $3 + 2(2) - 0 = 3 + 4 - 0 = 7 \neq 1$. This is false.
  • Therefore, plane (C) does not contain the line $L$.

• Option (D): $2x-y+z=5$

  • Check with $P_1(3,2,0)$: $2(3) - 2 + 0 = 6 - 2 + 0 = 4 \neq 5$. This is false.
  • Therefore, plane (D) does not contain the line $L$.

Step 5: Conclude the correct option. Only Option (A) satisfies the condition of containing the line $L$ (and thus the point $P_1$).

Common Mistakes & Tips

• Forgetting the initial check: Always verify if the given point lies on the given line. This single step can drastically simplify the problem, as it did here.
• Incomplete conditions for a plane containing a line: Remember that a plane contains a line if and only if it contains at least one point from the line AND its normal vector is perpendicular to the line's direction vector. Both conditions are necessary.
• Inefficient testing in MCQs: When checking options, substituting points from the line into the plane equations is usually the quickest method. If a single point (like $P_1$ here) satisfies multiple options, use a second point ($P_0$) or the normal vector perpendicularity check to narrow it down.
• If the point does NOT lie on the line: Do not get confused if the initial check fails. In that scenario, calculate the cross product of the vector connecting a point on the line to the external point, and the direction vector of the line. This cross product yields the normal vector of the plane.

Summary

The most critical first step in finding the equation of a plane that passes through a given point and contains a given line is to check if the point itself lies on the line. In this problem, the point $(3,2,0)$ was found to lie on the given line $\frac{x - 4}{1} = \frac{y - 7}{5} = \frac{z - 4}{4}$. This simplifies the task to finding a plane that merely contains the entire line. By testing the coordinates of two distinct points from the line (the given point $(3,2,0)$ and a point identified from the line's equation $(4,7,4)$) against the given options, we found that only option (A) $x-y+z=1$ satisfies the condition for both points. This confirms that the plane $x-y+z=1$ contains the given line and thus the given point.

The final answer is $\boxed{\text{A}}$