Question

Let the equation of the plane P containing the line \(x+10=\frac{8-y}{2}=z\) be \(ax + by + 3z = 2(a + b)\) and the distance of the plane P from the point (1, 27, 7) be c. Then a²+b²+c² is equal to ____________.

Answer 1

Correct Answer: 355

Step 1: Equation of the line. The given equation of the line is: \[ \frac{x + 10}{8} = \frac{-y}{2} = \frac{z}{3}. \] Let the common parameter be \( t \). Then the parametric equations of the line are: \[ x = 8t - 10, \quad y = -2t, \quad z = 3t. \] 

Step 2: Equation of the plane. The equation of the plane is given by: \[ ax + by + 3z = 2(a + b). \] Substituting the parametric equations of the line into the plane equation: \[ a(8t - 10) + b(-2t) + 3(3t) = 2(a + b). \] Simplify: \[ 8at - 10a - 2bt + 9t = 2a + 2b. \] Collect like terms: \[ (8a - 2b + 9)t = 10a + 2b. \] For the equation to hold true for all \( t \), the coefficients of \( t \) on both sides must be equal. Therefore: \[ 8a - 2b + 9 = 0 \quad \text{and} \quad 10a + 2b = 0. \] 

Step 3: Solving the system of equations. From the second equation: \[ 10a + 2b = 0 \quad \Rightarrow \quad 5a + b = 0 \quad \Rightarrow \quad b = -5a. \] Substitute this into the first equation: \[ 8a - 2(-5a) + 9 = 0 \quad \Rightarrow \quad 8a + 10a + 9 = 0 \quad \Rightarrow \quad 18a = -9 \quad \Rightarrow \quad a = -\frac{1}{2}. \] Then: \[ b = -5a = \frac{5}{2}. \] 

Step 4: Finding the distance of the point \( (1, 27, 7) \) from the plane. The formula for the distance from a point \( (x_1, y_1, z_1) \) to a plane \( ax + by + cz + d = 0 \) is: \[ d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}. \] The equation of the plane is: \[ -\frac{1}{2}x + \frac{5}{2}y + 3z - 5 = 0. \] Substitute \( (x_1, y_1, z_1) = (1, 27, 7) \) into the distance formula: \[ d = \frac{\left| -\frac{1}{2}(1) + \frac{5}{2}(27) + 3(7) - 5 \right|}{\sqrt{\left( -\frac{1}{2} \right)^2 + \left( \frac{5}{2} \right)^2 + 3^2}}. \] Simplify the numerator: \[ d = \frac{\left| -\frac{1}{2} + \frac{135}{2} + 21 - 5 \right|}{\sqrt{\frac{1}{4} + \frac{25}{4} + 9}} = \frac{\left| \frac{135 - 1}{2} + 16 \right|}{\sqrt{\frac{26}{4} + 9}}. \] \[ d = \frac{\left| \frac{134}{2} + 16 \right|}{\sqrt{\frac{26 + 36}{4}}} = \frac{75}{\sqrt{15}} = \frac{75}{\sqrt{15}}. \] Thus, \( c = \frac{75}{\sqrt{15}} \). 

Step 5: Calculating \( a^2 + b^2 + c^2 \). We already have: \[ a^2 = \left( -\frac{1}{2} \right)^2 = \frac{1}{4}, \quad b^2 = \left( \frac{5}{2} \right)^2 = \frac{25}{4}, \quad c^2 = \left( \frac{75}{\sqrt{15}} \right)^2 = \frac{5625}{15} = 375. \] Thus: \[ a^2 + b^2 + c^2 = \frac{1}{4} + \frac{25}{4} + 375 = \frac{26}{4} + 375 = 355. \]