<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] You can use Cauchy–Schwarz inequality to continue your approach.
As you wrote, we want to maximize
$$\frac{d^2}{a^2+b^2+c^2}=\frac{1}{\left(\frac ad\right)^2+\left(\frac bd\right)^2+\left(\frac cd\right)^2}$$(This is because we may suppose that $d\not=0$.)
In other words, we want to minimize
$$\left(\frac ad\right)^2+\left(\frac bd\right)^2+\left(\frac cd\right)^2.$$
Since the plane $ax+by+cz+d=0$ passes through $(4,3,5)$, we have
$$4a+3b+5c+d=0,$$
i.e.
$$4\cdot\frac ad+3\cdot\frac bd+5\cdot\frac cd=-1\tag1$$
By Cauchy–Schwarz inequality, 
$$\left(\left(\frac ad\right)^2+\left(\frac bd\right)^2+\left(\frac cd\right)^2\right)\left(4^2+3^2+5^2\right)\ge\left(4\cdot\frac ad+3\cdot\frac bd+5\cdot\frac cd\right)^2,$$
i.e.
$$\left(\frac ad\right)^2+\left(\frac bd\right)^2+\left(\frac cd\right)^2\ge\frac{\left(4\cdot\frac ad+3\cdot\frac bd+5\cdot\frac cd\right)^2}{4^2+3^2+5^2}=\frac{1}{50}.$$
The equality holds when
$$\frac{4d}{a}=\frac{3d}{b}=\frac{5d}{c}\tag2$$
From $(1)(2)$, the correct option is $(B)$.
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