<s>[INST] The equation of the plane which passes through the point of intersection of two space lines and at greatest distance from the point $(0,0,0)$ The equation of the plane which passes 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 the point $(0,0,0)$ is 
$(A)4x+3y+5z=25\hspace{1cm}(B)4x+3y+5z=50\hspace{1cm}(C)3x+4y+5z=49\hspace{1cm}(D)x+7y-5z=2$

I found the point of intersection of two lines as $(4,3,5)$.Let the equation of the required plane is $ax+by+cz+d=0$.Squared distance of plane $ax+by+cz+d=0$ from $(0,0,0)$ is $\frac{d^2}{a^2+b^2+c^2}$
We need to maximize $\frac{d^2}{a^2+b^2+c^2}$ under the constraint $4a+3b+5c+d=0$.
I cannot solve it further.I am stuck here.Please help me.Thanks.
 [/INST] An alternative to your approach is to this geometrically. To be the furthest distance then the plane must be perpendicular to the vector $(4,3,5)$. So to find the points of the plane $(x,y,z)$ we have:
$$(4,3,5)\cdot((x,y,z)-(4,3,5))=0$$
$$4x+3y+5x-16-9-25=0$$
$$4x+3y+5x=50$$
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