No Calculus, Think Differently ...!

\displaystyle{Let\quad Two\quad points\quad P\quad and\quad Q\quad in\quad co-ordinate\quad XYZ\quad Plane\\ such\quad that\quad \quad P(a,2-a,2a)\quad and\quad Q(b,b,3b)\\ Now\quad 'P'\quad Lies\quad on\quad Line\quad \frac { x }{ 1 } =\frac { y-2 }{ -1 } =\frac { z }{ 2 } \\ In\quad Vector\quad form\quad { eq }^{ n }\quad is:\\ \quad \xrightarrow { r } =\quad 2\overset { \^ }{ j } +\quad \lambda (\overset { \^ }{ i } -\overset { \^ }{ j } +2\overset { \^ }{ k } )\\ \\ Similarly\quad Q\quad lies\quad on\quad line\quad (I\quad write\quad Directly\quad Vector\quad form\quad of\quad { eq }^{ n })\quad is:\\ \xrightarrow { r } =\quad \mu (\overset { \^ }{ i } +\overset { \^ }{ j } +3\overset { \^ }{ k } )\\ \\ Now\quad To\quad minimize\quad 'E'\quad We\quad Have\quad to\quad find\quad out\\ Shortest\quad Distance\quad Between\quad The\quad two\quad given\quad lines\\ which\quad can\quad be\quad determined\quad easily.\\ \Longrightarrow \quad { E }_{ min }=\frac { 2 }{ \sqrt { 30 } } \\ so\quad m=2\quad \& \quad n=30} .

By QM-AM inequality,

$\sqrt {\frac {(0.2a-0.2b)^{2}+(0.2a-0.2b)^{2}+...+(0.2a-0.2b)^{2}+(2-a-b)^{2}+(1.5b-a)^{2}+(1.5b-a)^{2}+...+(1.5b-a)^{2}}{30}} \geq \frac {(0.2a-0.2b)+(0.2a-0.2b)+...+(0.2a-0.2b)+(2-a-b)+(1.5b-a)+(1.5b-a)+...+(1.5b-a)}{30}=\frac {2}{30}=\frac {1}{15}$

Here there are 25 terms of $(0.2a-0.2b)^{2}$ , 4 terms of $(1.5b-a)^{2}$ and one term of $(2-a-b)^{2}$ . Hence the answer is $\frac {2}{\sqrt {30}}$ . Equality can occur when $(a, b)=(\frac {17}{15}, \frac {4}{5})$

The QM-AM inequality is a form of Cauchy-Schwarz inequality. It states that the arithmetic mean of the squares in a set of real numbers, when square rooted, is greater than their arithmetic mean.

$Let\ x = a-b , y = a+b\\$ $E^2 = x^2 + (2-y)^2 + ( \frac{5}{2}x - \frac{1}{2}y)^2\\$ $E^2=\frac{29}{4}(x-\frac{5}{29}y)^2+\frac{30}{29}(y-\frac{29}{15})^2+\frac{2}{15}\\$ $E^2_{min} = \frac{2}{15}\\$ $Hence,\ E_{min} = \frac{2}{\sqrt{30}}\\$

E = f (a,b)

to find the minimum then f ' (a) = 0 & f ' (b) = 0

f ' (a) = 0

a - b = 1/3 ------------Eq (1)

f ' (b) = 0

11b - 6 a = 2 ------------Eq (2)

From 1 & 2

a = 17/15 , b = 12/15

Then E = 2/ sqrt (30)

m + n = 32

Let $\vec{a} = \langle 5, 1, - 2\rangle$ and $\vec{b} = \langle a-b, 2-a-b, 2a-3b\rangle$ . Note that $|\vec{a}| = \sqrt{30}$ and $|\vec{b}| = E$ . By Cauchy-Schwarz, we have $E\sqrt{30} = |\vec{a}||\vec{b}| \geq \vec{a} \cdot \vec{b} = 2$ . Therefore $E \geq \frac{2}{\sqrt{30}} \rightarrow \boxed{32}$ .

Just employ Cauchy Schwarz Inequality as $\left( (b-a)^2+(a+b-2)^2+(2a-3b)^2\right)\left(\frac {25}{4} +\frac {1}{4} +1\right)\ge \left(\frac {5(b-a)}{2}+\frac {a+b-2}{2}+(2a-3b)\right) ^2=1$

And hence follows the answer.