Key Concepts and Formulas

1. Equation of a Plane in Point-Normal Form: If a plane passes through a point $P_0(x_0, y_0, z_0)$ and has a normal vector $\vec{n} = A\hat{i} + B\hat{j} + C\hat{k}$, its equation is given by $A(x - x_0) + B(y - y_0) + C(z - z_0) = 0$.
2. Normal Vector from Foot of Perpendicular: If $P_0(x_0, y_0, z_0)$ is the foot of the perpendicular from the origin O(0,0,0) to a plane, then the vector $\vec{OP_0} = x_0\hat{i} + y_0\hat{j} + z_0\hat{k}$ is perpendicular to the plane. This makes $\vec{OP_0}$ a normal vector to the plane.
3. Volume of a Tetrahedron OABC: For a tetrahedron with one vertex at the origin O(0,0,0) and the other three vertices on the coordinate axes, say A($x_A$, 0, 0), B(0, $y_B$, 0), and C(0, 0, $z_C$), its volume is given by $V = \frac{1}{6} |x_A y_B z_C|$.

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Step-by-Step Solution

Step 1: Determine the Equation of Plane P We are given that the foot of the perpendicular from the origin O(0,0,0) to plane P is $P_0(2, a, 4)$.

• Why: According to the key concept, the vector from the origin to the foot of the perpendicular is a normal vector to the plane.
• The normal vector $\vec{n}$ to plane P is $\vec{OP_0} = (2-0)\hat{i} + (a-0)\hat{j} + (4-0)\hat{k} = 2\hat{i} + a\hat{j} + 4\hat{k}$.
• The plane P passes through the point $P_0(2, a, 4)$ and has the normal vector $\vec{n} = 2\hat{i} + a\hat{j} + 4\hat{k}$. Using the point-normal form of the plane equation: $2(x - 2) + a(y - a) + 4(z - 4) = 0$
• Simplifying the equation: $2x - 4 + ay - a^2 + 4z - 16 = 0$ $2x + ay + 4z = 20 + a^2 \quad \text{(Equation 1)}$

Step 2: Find the Intercepts of Plane P with the Coordinate Axes The plane P meets the coordinate axes at points A, B, and C.

• Why: To use the volume formula for the tetrahedron OABC, we need the coordinates of points A, B, and C, which are the intercepts of the plane with the x, y, and z axes, respectively.
• For point A (x-intercept): Set $y=0$ and $z=0$ in Equation 1. $2x + a(0) + 4(0) = 20 + a^2 \implies 2x = 20 + a^2 \implies x = \frac{20 + a^2}{2}$ So, $A \equiv \left(\frac{20 + a^2}{2}, 0, 0\right)$.
• For point B (y-intercept): Set $x=0$ and $z=0$ in Equation 1. $2(0) + ay + 4(0) = 20 + a^2 \implies ay = 20 + a^2 \implies y = \frac{20 + a^2}{a}$ So, $B \equiv \left(0, \frac{20 + a^2}{a}, 0\right)$.
• For point C (z-intercept): Set $x=0$ and $y=0$ in Equation 1. $2(0) + a(0) + 4z = 20 + a^2 \implies 4z = 20 + a^2 \implies z = \frac{20 + a^2}{4}$ So, $C \equiv \left(0, 0, \frac{20 + a^2}{4}\right)$.

Step 3: Calculate the Volume of Tetrahedron OABC We are given that the volume of the tetrahedron OABC is 144 unit$^3$.

• Why: We will use the given volume and the intercepts found in Step 2 to form an equation involving 'a', which we can then solve.
• Using the formula $V = \frac{1}{6} |x_A y_B z_C|$: $V = \frac{1}{6} \left| \left(\frac{20 + a^2}{2}\right) \left(\frac{20 + a^2}{a}\right) \left(\frac{20 + a^2}{4}\right) \right|$ Since $a \in N$ (natural numbers), $a > 0$, and $20+a^2 > 0$, the absolute value is not needed. $V = \frac{1}{6} \frac{(20 + a^2)^3}{8a}$
• Equating this to the given volume, 144: $144 = \frac{(20 + a^2)^3}{48a}$

Step 4: Solve for 'a'

• Why: We need to find the specific value of 'a' to determine the exact equation of the plane.
• Rearranging the equation from Step 3: $(20 + a^2)^3 = 144 \times 48a$ $(20 + a^2)^3 = 6912a \quad \text{(Equation 2)}$
• Since $a \in N$, we can test small natural numbers for $a$:
  • If $a = 1$: LHS = $(20 + 1^2)^3 = (21)^3 = 9261$. RHS = $6912 \times 1 = 6912$. LHS $\neq$ RHS.
  • If $a = 2$: LHS = $(20 + 2^2)^3 = (20 + 4)^3 = (24)^3 = 13824$. RHS = $6912 \times 2 = 13824$. LHS = RHS.
• Thus, $a=2$ is the correct value.

Step 5: Determine the Final Equation of Plane P

• Why: With the value of 'a' determined, we can now write the concrete equation for plane P.
• Substitute $a=2$ back into Equation 1: $2x + (2)y + 4z = 20 + (2)^2$ $2x + 2y + 4z = 20 + 4$ $2x + 2y + 4z = 24$
• Divide the entire equation by 2 to simplify: $x + y + 2z = 12 \quad \text{(Equation of Plane P)}$

Step 6: Check Which Point is NOT on Plane P

• Why: The question asks which of the given points is NOT on the plane. We check each option by substituting its coordinates into the plane equation $x + y + 2z = 12$.
• (A) $(3,0,4)$: Substitute $x=3, y=0, z=4$: $3 + 0 + 2(4) = 3 + 0 + 8 = 11$ Since $11 \neq 12$, this point does NOT lie on the plane.
• (B) $(0,6,3)$: Substitute $x=0, y=6, z=3$: $0 + 6 + 2(3) = 0 + 6 + 6 = 12$ Since $12 = 12$, this point lies on the plane.
• (C) $(0,4,4)$: Substitute $x=0, y=4, z=4$: $0 + 4 + 2(4) = 0 + 4 + 8 = 12$ Since $12 = 12$, this point lies on the plane.
• (D) $(2,2,4)$: Substitute $x=2, y=2, z=4$: $2 + 2 + 2(4) = 2 + 2 + 8 = 12$ Since $12 = 12$, this point lies on the plane.

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Common Mistakes & Tips

• Misinterpreting "Foot of Perpendicular": A common error is not realizing that the vector from the origin to the foot of the perpendicular is directly the normal vector to the plane.
• Volume Formula for Tetrahedron: Ensure you use the correct formula. For a tetrahedron with vertices at the origin and on the axes, $V = \frac{1}{6} |x_A y_B z_C|$ is a powerful shortcut. For general vertices, a more complex determinant formula involving vectors would be needed.
• Algebraic Precision: Be careful with calculations, especially when dealing with powers and products of expressions involving 'a'. Testing integer values for 'a' when it's a natural number can save time compared to attempting to solve a cubic equation algebraically.

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Summary

The problem required us to find the equation of a plane given the foot of the perpendicular from the origin, and then use the volume of the tetrahedron formed by the plane's intercepts with the axes and the origin to determine an unknown parameter 'a'. Once 'a' was found, the plane's equation was finalized, and each given point was tested to identify which one does not lie on the plane. The key steps involved identifying the normal vector from the foot of the perpendicular, setting up the plane equation, calculating axis intercepts, using the tetrahedron volume formula to solve for 'a', and finally verifying the options.

The final answer is $\boxed{(3,0,4)}$ which corresponds to option (A).