Question

If equation of plane passing through the point of intersection of lines $\frac{x-1}{3}=\frac{y-2}{1}=\frac{z-3}{2}$ and $\frac{x-3}{1}=\frac{y-1}{2}=\frac{z-2}{3}$ and at greatest distance from origin is $4x+ay+bz=c$, then find the value of $(a+b+c)$

Answer 1

Answer: 58

Point of intersection of the given lines is (4, 3, 5).
Required plane is through $(4,3,5)$ and perpendicular to line joing $(4,3,5)\&(0,0,0)$.
$\Rightarrow$ Equation of planes $4(x-4)+3(y-3)+5(z-5)=0$