The foot of perpendicular of the point 2 , 0 , 5 on the line x + 1 2 = y - 1 5 = z + 1 - 1 is α , &#946…

Given line is

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

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

$\overrightarrow{PA}=\left(3-2\lambda \right)\hat{i}-\left(5\lambda +1\right)\hat{j}+\left(6+\lambda \right)\hat{k}$

Direction ratio of line is

$\overrightarrow{b}=2\hat{i}+5\hat{j}-\hat{k}$

Now, $\overrightarrow{PA}\perp L$, so

$\overrightarrow{PA}·\overrightarrow{b}=0$

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

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

$\Rightarrow -30\lambda -5=0$

$\Rightarrow \lambda =-\frac{1}{6}$

So,

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

Now,

$\alpha =2\left(-\frac{1}{6}\right)-1=-\frac{4}{3}\Rightarrow \boxed{\alpha =-\frac{4}{3}}$

$\beta =5\left(-\frac{1}{6}\right)+1=\frac{1}{6}\Rightarrow \boxed{\beta =\frac{1}{6}}$

$\gamma =-\lambda -1=\frac{1}{6}-1\Rightarrow \boxed{\gamma =-\frac{5}{6}}$

So,

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