The equation of the plane which passes through the point of intersection of two space lines and at greatest distance from the point $(0,0,0)$

Question

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
$(A)4x+3y+5z=25\hspace{1cm}(B)4x+3y+5z=50\hspace{1cm}(C)3x+4y+5z=49\hspace{1cm}(D)x+7y-5z=2$

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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.

Answer 1

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:

$$(4,3,5)\cdot((x,y,z)-(4,3,5))=0$$ $$4x+3y+5x-16-9-25=0$$ $$4x+3y+5x=50$$

Answer 2

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 $(B)$.