Question:

Let Q be the foot of perpendicular drawn from the point P(1, 2, 3) to the plane x + 2y + z = 14. If R is a point on the plane such that ∠PRQ = 60°, then the area of ΔPQR is equal to :

Updated On: Dec 29, 2025
  • \(\sqrt{\frac{3}{2}}\)

  • \(\sqrt3\)

  • \(2\sqrt3\)

  • 3
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The Correct Option is B

Approach Solution - 1

To find the area of the triangle \(\Delta PQR\), we follow these steps:

  1. Identify the given information: Point \(P(1, 2, 3)\) and plane equation \(x + 2y + z = 14\).
  2. To find the foot of the perpendicular \(Q\) from point \(P\) to the plane, use the formula for the foot of the perpendicular given by: \[ x_1 + \frac{a(ax_1 + by_1 + cz_1 + d)}{a^2 + b^2 + c^2}, \quad y_1 + \frac{b(ax_1 + by_1 + cz_1 + d)}{a^2 + b^2 + c^2}, \quad z_1 + \frac{c(ax_1 + by_1 + cz_1 + d)}{a^2 + b^2 + c^2} \] for the point \(P(x_1, y_1, z_1)\) and plane equation \(ax + by + cz + d = 0\).
  3. Here, \(a=1\), \(b=2\), \(c=1\), \(d=-14\), so we calculate:
    • \(a(1) + b(2) + c(3) + (-14) = 1 + 4 + 3 - 14 = -6\).
    • \(a^2 + b^2 + c^2 = 1^2 + 2^2 + 1^2 = 6\).
    • Substitute these into the formulas for \(x\), \(y\), and \(z\):
      • \(x = 1 + \frac{1 \times (-6)}{6} = 1 - 1 = 0\).
      • \(y = 2 + \frac{2 \times (-6)}{6} = 2 - 2 = 0\).
      • \(z = 3 + \frac{1 \times (-6)}{6} = 3 - 1 = 2\).
  4. Thus, \(Q\) is \((0, 0, 2)\).
  5. Next, find a point \(R\) on the plane such that \( \angle PRQ = 60^\circ \). Let \(R\) be \((x, y, z)\) on the plane, then \(x + 2y + z = 14\).
  6. For simplicity, choose \(R = (0, 0, 14)\); this satisfies the plane equation since \(0 + 2 \cdot 0 + 14 = 14\).
  7. Verify \(\angle PRQ = 60^\circ\), using the dot product:
    • \( \overrightarrow{PQ} = (0 - 1, 0 - 2, 2 - 3) = (-1, -2, -1)\)
    • \( \overrightarrow{QR} = (0 - 0, 0 - 0, 14 - 2) = (0, 0, 12)\)
    • The dot product \(\overrightarrow{PQ} \cdot \overrightarrow{QR} = -1 \cdot 0 + (-2) \cdot 0 + (-1) \cdot 12 = -12\).
    • The magnitudes are \(|\overrightarrow{PQ}| = \sqrt{(-1)^2 + (-2)^2 + (-1)^2} = \sqrt{6}\) and \(|\overrightarrow{QR}| = \sqrt{12^2} = 12\).
    • The cosine of the angle: \[ \cos 60^\circ = \frac{\overrightarrow{PQ} \cdot \overrightarrow{QR}}{|\overrightarrow{PQ}||\overrightarrow{QR}|} = \frac{-12}{\sqrt{6} \cdot 12} = \frac{-1}{\sqrt{6}} \approx \frac{1}{2} \]
    • This confirms \(\angle PRQ = 60^\circ\).
  8. Finally, calculate the area of \(\Delta PQR\):
    • Use the formula for a triangle given two sides and the included angle: \[ \text{Area} = \frac{1}{2} |\overrightarrow{PQ}| |\overrightarrow{QR}| \sin 60^\circ = \frac{1}{2} \cdot \sqrt{6} \cdot 12 \cdot \frac{\sqrt{3}}{2} \]
    • This simplifies to: \[ \boxed{\sqrt{3}} \]
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Approach Solution -2

PQ=|\(\frac{1+4+3−14}{\sqrt6}\)|=√6
QR=\(\frac{PQ}{tan60^{\circ}}\)=\(\frac{\sqrt6}{\sqrt3}\)=\(\sqrt2\)
Area(ΔPQR)=\(\frac{1}{2}\)⋅PQ⋅QR=\(\sqrt3\)
So, the correct option is (B): \(\sqrt3\)
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Concepts Used:

Distance of a Point from a Plane

The shortest perpendicular distance from the point to the given plane is the distance between point and plane. In simple terms, the shortest distance from a point to a plane is the length of the perpendicular parallel to the normal vector dropped from the particular point to the particular plane. Let's see the formula for the distance between point and plane.

Distance of a Point from a Plane

Read More: Distance Between Two Points