If a triangle lies on A(2,-1,0), B(1,2,3), C(-1,0,1) with its orthocenter being H(a,b,c), what is a+b+c?

Here's my algebraic approach, though it's even longer than heureka's!

.

Here's a Geogebra visual solution (sorry, I used D for the orthocenter, not H):

.

Thanks Alan,

I've never tried to draw anything so complicated.  

But, I would like to see someone do this algebraically. 

I don't have time and to be honest I don't think I know how.

Do you need to find 3 dimensional gradients ???

Let triangle ABC be A(2,-1,0), B(1,2,3), C(-1,0,1), with its orthocenter being H(a,b,c), what is a, b, and c ?

$$\\\small{\text{
$
\vec{u}=\vec{a}-\vec{c}=
\left(
\begin{array}{c}2-(-1)\\-1-0 \\0-1 \end{array}
\right)=\left(
\begin{array}{c}3\\-1 \\-1 \end{array}
\right)\qquad
\vec{v}=\vec{b}-\vec{c}=
\left(
\begin{array}{c}1-(-1)\\2-0 \\3-1 \end{array}
\right)=\left(
\begin{array}{c}2\\2 \\2\end{array}
\right)
$
}}\\\\
\small{\text{
$
u=|\vec{u}|=\sqrt{3^2+(-1)^2+(-1)^2}=\sqrt{11} \qquad
v=|\vec{v}|=\sqrt{2^2+2^2+2^2}=\sqrt{12}
$
}}\\\\
\small{\text{
$
\vec{u_0}=\dfrac{\vec{u}}{u}=
\frac{1}{\sqrt{11}}
\left(
\begin{array}{c}3\\-1 \\-1 \end{array}
\right)\qquad
\vec{v_0}=\dfrac{\vec{v}}{v}=
\frac{1}{\sqrt{12}}
\left(
\begin{array}{c}2\\2 \\2 \end{array}
\right)\qquad
$
}}\\\\
\boxed{
\small{\text{
$
\vec{H}=(\vec{u_0}\cdot \vec{v}) \cdot \vec{u_0}+ \mu [\vec{v} - (\vec{u_0}\cdot \vec{v}) \cdot \vec{u_0} ] + \vec{c}
$
}}
}\\\\
\boxed{
\small{\text{
$
\vec{H}=(\vec{v_0}\cdot \vec{u}) \cdot \vec{v_0}+ \lambda [\vec{u} - (\vec{v_0}\cdot \vec{u}) \cdot \vec{v_0} ] + \vec{c}
$
}}
}\\\\
\boxed{
\small{\text{
$
\vec{H}-\vec{c}=
(\vec{u_0}\cdot \vec{v}) \cdot \vec{u_0}+ \mu [\vec{v} - (\vec{u_0}\cdot \vec{v}) \cdot \vec{u_0} ]
=
(\vec{v_0}\cdot \vec{u}) \cdot \vec{v_0}+ \lambda [\vec{u} - (\vec{v_0}\cdot \vec{u}) \cdot \vec{v_0} ]
$
}}
}\\\\
\small{\text{
$
\mu [\vec{v} - (\vec{u_0}\cdot \vec{v}) \cdot \vec{u_0} ]
-\lambda [\vec{u} - (\vec{v_0}\cdot \vec{u}) \cdot \vec{v_0} ]
=(\vec{v_0}\cdot \vec{u}) \cdot \vec{v_0}
-(\vec{u_0}\cdot \vec{v}) \cdot \vec{u_0} \quad | \quad \cdot \vec{v_0}
$
}}\\\\
\small{\text{
$
\mu [\vec{v} - (\vec{u_0}\cdot \vec{v}) \cdot \vec{u_0} ]\cdot \vec{v_0}
-\underbrace{\lambda [\vec{u} - (\vec{v_0}\cdot \vec{u}) \cdot \vec{v_0} ]\cdot \vec{v_0}}_{=0}
=(\vec{v_0}\cdot \vec{u}) \cdot (\vec{v_0}\cdot \vec{v_0} )
-(\vec{u_0}\cdot \vec{v}) \cdot (\vec{u_0}\cdot \vec{v_0} )
$
}}\\\\
\small{\text{
$
\mu [\vec{v} - (\vec{u_0}\cdot \vec{v}) \cdot \vec{u_0} ]\cdot \vec{v_0}
=(\vec{v_0}\cdot \vec{u}) \cdot \underbrace{(\vec{v_0}\cdot \vec{v_0} )}_{=1}
-(\vec{u_0}\cdot \vec{v}) \cdot (\vec{u_0}\cdot \vec{v_0} )
$
}}\\\\
\small{\text{
$
\mu [\underbrace{(\vec{v}\cdot \vec{v_0})}_{=v} - (\vec{u_0}\cdot \vec{v}) \cdot (\vec{u_0}\cdot \vec{v_0}) ]
=(\vec{v_0}\cdot \vec{u})
-(\vec{u_0}\cdot \vec{v}) \cdot (\vec{u_0}\cdot \vec{v_0} )
$
}}\\\\
\small{\text{
$
\mu [v - (\vec{u_0}\cdot \vec{v}) \cdot (\vec{u_0}\cdot \vec{v_0}) ]
=(\vec{v_0}\cdot \vec{u})
-(\vec{u_0}\cdot \vec{v}) \cdot (\vec{u_0}\cdot \vec{v_0} )
$
}}$$

$$\boxed{
\small{\text{
$
\mu=
\dfrac{
(\vec{v_0}\cdot \vec{u})
-(\vec{u_0}\cdot \vec{v}) \cdot (\vec{u_0}\cdot \vec{v_0} )
}
{v - (\vec{u_0}\cdot \vec{v}) \cdot (\vec{u_0}\cdot \vec{v_0}) }
\qquad
\vec{H}=(\vec{u_0}\cdot \vec{v}) \cdot \vec{u_0}+ \mu [\vec{v} - (\vec{u_0}\cdot \vec{v}) \cdot \vec{u_0} ] + \vec{c}
$
}}
}\\\\
\small{\text{
$
\vec{v}\cdot \vec{u} = \left(\begin{array}{c} 2\\2 \\2 \end{array}\right)
\cdot \left(\begin{array}{c} 3\\-1 \\-1 \end{array}\right) =6-2-2 = 2
$
}}\\\\
\small{\text{
$
\vec{v_0}\cdot \vec{u} = \dfrac{\vec{v}\cdot \vec{u}}{v} = \dfrac{2}{\sqrt{12}}
\qquad \vec{u_0}\cdot \vec{v} = \dfrac{\vec{v}\cdot \vec{u}}{u} = \dfrac{2}{\sqrt{11}}
\qquad \vec{u_0}\cdot \vec{v_0} = \dfrac{\vec{v}\cdot \vec{u}}{v\cdot u} = \dfrac{2}{\sqrt{11}\cdot \sqrt{12}}
$
}}\\\\
\small{\text{
$
\mu=
\dfrac{
\dfrac{2}{\sqrt{12}}
-\dfrac{2}{\sqrt{11}} \cdot \dfrac{2}{\sqrt{11}\cdot \sqrt{12}}
}
{\sqrt{12} - \dfrac{2}{\sqrt{11}} \cdot \dfrac{2}{\sqrt{11}\cdot \sqrt{12}} }
=\dfrac{
\dfrac{1}{\sqrt{12}}\cdot \left(2-\dfrac{4}{11} \right) }
{\dfrac{1}{\sqrt{12}}\cdot \left( 12 -\dfrac{4}{11} \right) }
= \dfrac{18}{128} = \dfrac{9}{64}
$
}}$$

$$\\\small{\text{
$
\vec{H}=(\vec{u_0}\cdot \vec{v}) \cdot \vec{u_0}+ \mu [\vec{v} - (\vec{u_0}\cdot \vec{v}) \cdot \vec{u_0} ] + \vec{c}
$
}}\\
\small{\text{
$
\vec{H}=(1-\mu )[(\vec{u_0}\cdot \vec{v}) \cdot \vec{u_0} ]+\mu\vec{v}+ \vec{c}
$
}}\\\\
\small{\text{
$
\vec{H}=
\left(1-\dfrac{9}{64} \right)\cdot
\left[ \dfrac{2} { \sqrt{11} } \cdot \vec{u_0} \right]
+ \dfrac{9}{64}\cdot \vec{v}
+ \vec{c}
$
}}\\\\
\small{\text{
$
\vec{H}=
\left(\dfrac{55}{64} \right)\cdot \left( \dfrac{2} { \sqrt{11} }\right) \cdot \dfrac{\vec{u}}{u}
+ \dfrac{9}{64}\cdot \vec{v}
+ \vec{c}
$
}}\\\\
\small{\text{
$
\vec{H}=
\left(\dfrac{55}{64} \right)\cdot \left( \dfrac{2} { \sqrt{11} }\right) \cdot \dfrac{\vec{u}}{\sqrt{11}}
+ \dfrac{9}{64}\cdot \vec{v}
+ \vec{c}
$
}}\\\\
\small{\text{
$
\vec{H}=
\dfrac{10}{64}\cdot \vec{u}+ \dfrac{9}{64}\cdot \vec{v} + \vec{c}
$
}}$$

$$\\ \small{\text{
$
\vec{H}=
\dfrac{10}{64}\cdot
\left(\begin{array}{c}3 \\ -1\\ -1\end{array}\right)
+ \dfrac{9}{64}\cdot
\left(\begin{array}{c}2 \\ 2\\ 2\end{array} \right)
+
\left( \begin{array}{c}-1 \\ 0\\ 1\end{array} \right)
$
}}\\\\\\
\small{\text{
$
\vec{H}=
\left(\begin{array}{c}\dfrac{30}{64}+\dfrac{18}{64}-1 \\ \\ -\dfrac{10}{64}+\dfrac{18}{64}+0\\ \\ -\dfrac{10}{64}+\dfrac{18}{64}+1\end{array}\right)
$
}}\\\\\\
\small{\text{
$
\vec{H}=
\left(\begin{array}{c} -\dfrac{1}{4} \\\\ \dfrac{1}{8} \\\\ 1 \dfrac{1}{8}
\end{array}
\right)
$
}}\\\\\\
\small{\text{
$
\vec{H}=
\left(\begin{array}{c} -0.25 \\ 0.125 \\ 1.125
\end{array}
\right)
$
}}$$

Thank you.      That is a masterpeice Heureka.   

Unfortunately I do not understand vectors very well so I got lost fairly quickly.

I will try to take time out to understand it better. 

I should  do some study on vectors.  There are just so many things to learn.  

Thanks Alan for that great matrix answer.  

I need to have a proper look at all of these solutions.

We have not heard from the anon that asked this question.  I wonder which answer they were after?