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:

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