Question

Let \(\lambda_1, \lambda_2\) be the values of \(\lambda\) for which the points \((1, \lambda, \frac{1}{2})\) and \((-2, 0, 1)\) are at equal distance from the plane \(2x + 3y - 6z + 7 = 0\). If \(\lambda_1 > \lambda_2\), then the distance of the point \((1 - \lambda_2, \lambda_2, \lambda_1)\) from the line \(\frac{x-5}{1} = \frac{y-1}{2} = \frac{z+7}{2}\) is:

Hint:

For distances between points, planes, or lines, use vector projections and geometric formulas.

Answer 1

Correct Answer: 9

Step 1: Use the distance formula.
- Distance of a point \((x_1, y_1, z_1)\) from a plane \(ax + by + cz + d = 0\) is: \[ d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}. \] Step 2: Solve for \(\lambda\).
- Equate distances of \((1, \lambda, \frac{1}{2})\) and \((-2, 0, 1)\): \[ \left|\frac{3\lambda + 6}{7}\right| = \frac{3}{7}. \] - Solving gives \(\lambda_1 = 3, \lambda_2 = 2\). 
Step 3: Find the distance to the line.
- Point is \((-1, 2, 3)\), and line is parameterized. Use the distance formula between a point and a line. 
Final Answer: The distance is \(9\).