Question

The ratio in which the surface $x^2+y^2+z^2=25$ divides the line joining $(0,1,2)$ and $(3,4,5)$ is $\frac{a\pm \sqrt{b}}{c}$ then $a+b+c=$

Answer 1

Let the line segment has a point $R$ intersecting on the surface of the sphere
$\Rightarrow R=(\frac{3\lambda }{1+\lambda }+\frac{4\lambda +1}{1+\lambda },\frac{5\lambda +2}{1+\lambda })$
So ${9\lambda }^2+{16\lambda }^2+1+8\lambda +{25\lambda }^2+4+20\lambda -25\{{\lambda }^2+1+2\lambda \}=0$
$\Rightarrow {25\lambda }^2-22\lambda -20=0$
$\Rightarrow \lambda =\frac{22\pm \sqrt{484+2000}}{50}=\frac{22\pm \sqrt{2484}}{50}=\frac{11\pm \sqrt{621}}{25}$

Then, the value of $a+b+c=11+621+25=657$.