Question

Let a line passing through the point \((-1, 2, 3)\) intersect the lines\( L_1 : \frac{x - 1}{3} = \frac{y - 2}{2} = \frac{z + 1}{-2} \) at \( M(\alpha, \beta, \gamma) \) and\( L_2 : \frac{x + 2}{-3} = \frac{y - 2}{-2} = \frac{z - 1}{4} \) at \( N(a, b, c) \).Then the value of \( \frac{(\alpha + \beta + \gamma)^2}{(a + b + c)^2} \) equals ____.

Answer 1

Correct Answer: 196

Write the equation of the line passing through \( (-1, 2, 3) \) with direction ratios \( (l, m, n) \):

\[ x = -1 + \lambda l, \quad y = 2 + \lambda m, \quad z = 3 + \lambda n. \]

Intersection with \( L_1 \): For intersection, equate:

\[ -1 + \lambda l = 1 + 3\mu, \quad 2 + \lambda m = 2 + 2\mu, \quad 3 + \lambda n = -1 - 2\mu. \]

Intersection with \( L_2 \): For intersection, equate:

\[ -1 + \lambda l = -2 - 3\nu, \quad 2 + \lambda m = 2 + 4\nu, \quad 3 + \lambda n = 1 - 2\nu. \]

Solve for \( \alpha, \beta, \gamma, a, b, \) and \( c \).

Calculate:

\[ \frac{(\alpha + \beta + \gamma)^2}{a + b + c} = 196. \]