Question

Let the plane P pass through the intersection of the planes \(2 x+3 y-z=2\)and \(x+2 y+3 z=6,\) and be perpendicular to the plane \(2 x+y-z+1=0\)If d is the distance of P from the point (-7,1,1), then \(d^2\) is equal to :

• A. \(\frac{25}{83}\)
• B. \(\frac{250}{83}\)
• C. \(\frac{15}{53}\)
• D. \(\frac{250}{82}\)

Hint:

For planes passing through the intersection of two planes, use the family of planes formula. To ensure perpendicularity, use the dot product of normal vectors.

Answer 1

The Correct Option is B

The plane P passes through the intersection of the planes: \[ P_1: 2x + 3y - z - 2 = 0 \quad \text{and} \quad P_2: x + 2y + 3z - 6 = 0. \] The equation of P is: \[ P \equiv P_1 + \lambda P_2 = 0. \] Substituting the equations of \(P_1 \ and \  P_2\): \[ (2 + \lambda)x + (3 + 2\lambda)y + (-1 + 3\lambda)z - (2 + 6\lambda) = 0. \tag{1} \] Step 2: Determine λ  Since P is perpendicular to the plane \(P_3\): 2x + y - z + 1 = 0, the normal vector of P is perpendicular to the normal vector of P_3. This gives: \[ \vec{n}_P \cdot \vec{n}_3 = 0, \] where \(\vec{n}_P = (2 + \lambda, 3 + 2\lambda, -1 + 3\lambda)  \ and \  \vec{n}_3 = (2, 1, -1).\) Taking the dot product: \[ (2 + \lambda)(2) + (3 + 2\lambda)(1) + (-1 + 3\lambda)(-1) = 0. \] Simplify: \[ 4 + 2\lambda + 3 + 2\lambda + 1 - 3\lambda = 0, \quad 8 + \lambda = 0. \] Solve for \lambda: \[ \lambda = -8. \tag{2} \] Step 3: Equation of Plane P  Substitute \lambda = -8 into equation (1): \[ P = (-6)x - (13)y + (-25)z + 46 = 0. \tag{3} \] Step 4: Distance of Point (-7, 1, 1) from Plane P  The distance d of a point (x_1, y_1, z_1) from a plane ax + by + cz + d = 0 is: \[ d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}. \] Here, (x_1, y_1, z_1) = (-7, 1, 1), and the equation of P is -6x - 13y - 25z + 46 = 0. Substituting: \[ d = \frac{|(-6)(-7) + (-13)(1) + (-25)(1) + 46|}{\sqrt{(-6)^2 + (-13)^2 + (-25)^2}}. \] Simplify: \[ d = \frac{|42 - 13 - 25 + 46|}{\sqrt{36 + 169 + 625}}, \] \[ d = \frac{50}{\sqrt{830}}. \] Step 5: Calculate d^2 \[ d^2 = \left(\frac{50}{\sqrt{830}}\right)^2 = \frac{2500}{830} = \frac{250}{83}. \] Conclusion:  \[ d^2 = \frac{250}{83} \quad . \]