Chọn C

Ta có \(d\left( A,\left( P \right) \right)\)\( =\frac{\left| 2+m+6m+3-m-2 \right|}{\sqrt{{{1}^{2}}+{{m}^{2}}+{{\left( 2m+1 \right)}^{2}}}}\)\( =\sqrt{\frac{{{\left( 6m+3 \right)}^{2}}}{5{{m}^{2}}+4m+2}}\)

Xét \(f\left( m \right)\)\( =\frac{{{\left( 6m+3 \right)}^{2}}}{5{{m}^{2}}+4m+2}\)\( \Rightarrow {f}'\left( m \right)\)\( =\frac{\left( 6m+3 \right)\left( -6m+12 \right)}{{{\left( 5{{m}^{2}}+4m+2 \right)}^{2}}}=0\)

\(\Leftrightarrow \left[ \begin{align} & m=\frac{-1}{2} \\ & m=2 \\ \end{align} \right.\)

Vậy \(\max \ d\left( A,\left( P \right) \right)=\frac{\sqrt{30}}{2}\) khi và chỉ khi \(m=2\).

Vậy \(\left( P \right):x+2y+5z-4=0\).

Gọi \(d\) là đường thẳng qua \(A\) và vuông góc \(\left( P \right)\) nên \(d:\left\{ \begin{align} & x=2+t \\ & y=1+2t \\ & z=3+5t \\ \end{align} \right.\)

Khi đó \(H=d\cap \left( P \right)\)

Ta có \(2+t+2+4t+15+25t-4=0\Leftrightarrow 30t=-15\Leftrightarrow t=\frac{-1}{2}\)

\(H\left( \frac{3}{2};0;\frac{1}{2} \right)\). Do đó \(a+b=\frac{3}{2}\)

Cách 2: Khi đó \(M\left( x;y;z \right)\)cố định của \(\left( P \right)\) thỏa \(m\left( y+2z-1 \right)+x+z-2=0\quad \forall m\)

\(\begin{array}{l} \Leftrightarrow \left\{ \begin{array}{l} y + 2z - 1 = 0\\ x + z - 2 = 0 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} x = 2 - t\\ y = 1 - 2t\\ z = t \end{array} \right.\quad \left( \Delta \right) \end{array}\)

\(\left( P \right)\) luôn qua \(\left( \Delta  \right)\) cố định

\(AH\le AK\)\( \Rightarrow d{{\left( A;\left( P \right) \right)}_{\max }}\)\( \Rightarrow H\equiv K\)

\(K\left( 2-t;1-2t;t \right)\) vì qua \(\overrightarrow{AK}.\overrightarrow{{{u}_{\Delta }}}=0\)\( \Rightarrow t=\frac{1}{2}\)\( \Rightarrow K\left( \frac{3}{2};0;\frac{1}{2} \right)\)

Suy ra \(H\equiv K\) nên \(H\left( \frac{3}{2};0;\frac{1}{2} \right)\)