The foot of perpendicular of the point (2, 0, 5) on the line (x +...

Given line is

$L: \dfrac{x + 1}{2} = \dfrac{y - 1}{5} = \dfrac{z + 1}{- 1} = \lambda$ (let)

Let foot of perpendicular is $P\left(2\lambda - 1, 5\lambda + 1, - \lambda - 1\right)$

\overset{\to}{PA} = \left(3 - 2\lambda\right)\overset{^}{i} - \left(5\lambda + 1\right)\overset{^}{j} + \left(6 + \lambda\right)\overset{^}{k}

Direction ratio of line is

\overset{\to}{b} = 2\overset{^}{i} + 5\overset{^}{j} - \overset{^}{k}

Now, $\overset{\to}{PA} ⊥ L$, so

$\overset{\to}{PA} \cdot \overset{\to}{b} = 0$

$⇒ 2\left(3 - 2\lambda\right) - 5\left(5\lambda + 1\right) - \left(6 + \lambda\right) = 0$

$⇒ 6 - 4\lambda - 25\lambda - 5 - 6 - \lambda = 0$

$⇒ - 30\lambda - 5 = 0$

$⇒ \lambda = - \dfrac{1}{6}$

So,

$P\left(2\lambda - 1, 5\lambda + 1, - \lambda - 1\right) ≡ P\left(\alpha, \beta, \gamma\right)$

Now,

$\alpha = 2\left(- \dfrac{1}{6}\right) - 1 = - \dfrac{4}{3} ⇒ \alpha = - \dfrac{4}{3}$

$\beta = 5\left(- \dfrac{1}{6}\right) + 1 = \dfrac{1}{6} ⇒ \beta = \dfrac{1}{6}$

$\gamma = - \lambda - 1 = \dfrac{1}{6} - 1 ⇒ \gamma = - \dfrac{5}{6}$

So,

$\dfrac{\beta}{\gamma} = - \dfrac{1}{5} \ne - 5$