Reflection of a point across the plane $5 x - 3 y + 2 z = 10$

Let $\mathbf{p} = (p_1, p_2, p_3)$ and $\mathbf{p'} = (p_1', p_2', p_3')$ .

The midpoint of $\mathbf{p}$ and $\mathbf{p'}$ lies on the plane, so $5\bigg(\cfrac{p_1 + p_1'}{2}\bigg) - 3\bigg(\cfrac{p_2 + p_2'}{2}\bigg) + 2\bigg(\cfrac{p_3 + p_3'}{2}\bigg) = 10$ , or:

$5(p_1 + p_1') - 3(p_2 + p_2') + 2(p_3 + p_3') = 20$

$\overrightarrow{\mathbf{p'p}}$ is the same direction as the normal of the equation, which is $(5, -3, 2)$ , so $p_1 - p_1' = 5k$ , $p_2 - p_2' = -3k$ , and $p_3 - p_3' = 2k$ for some constant $k$ , so $5(p_1 - p_1') - 3(p_2 - p_2') + 2(p_3 - p_3') = 25k + 9k + 4k$ , or:

$5(p_1 - p_1') - 3(p_2 - p_2') + 2(p_3 - p_3') = 38k$

Adding these two equations gives $10p_1 - 6p_2 + 4p_3 = 20 + 38k$ , which rearranges to $k = \frac{1}{19}(5p_1 - 3p_2 + 2p_3 - 10)$ .

Then:

$p_1 - p_1' = 5k$ rearranges to $p_1' = p_1 - 5k = p_1 - 5 \cdot \frac{1}{19}(5p_1 - 3p_2 + 2p_3 - 10)$ or $p_1' = \frac{1}{19}(-6p_1 + 15p_2 - 10p_3) + \frac{1}{19}(50)$

$p_2 - p_2' = -3k$ rearranges to $p_2' = p_2 - 5k = p_2 + 3 \cdot \frac{1}{19}(5p_1 - 3p_2 + 2p_3 - 10)$ or $p_2' = \frac{1}{19}(15p_1 + 10p_2 + 6p_3) + \frac{1}{19}(-30)$

$p_3 - p_3' = 2k$ rearranges to $p_3' = p_3 - 5k = p_3 - 2 \cdot \frac{1}{19}(5p_1 - 3p_2 + 2p_3 - 10)$ or $p_3' = \frac{1}{19}(-10p_1 + 6p_2 + 15p_3) + \frac{1}{19}(20)$

which means:

$\mathbf{p'} = \cfrac{1}{19} \begin{bmatrix} -6 & 15 & -10 \\ 15 & 10 & 6 \\ -10 & 6 & 15 \end{bmatrix} \mathbf{p} + \cfrac{1}{19} \begin{bmatrix} 50 \\ -30 \\ 20 \end{bmatrix}$

so that $a = -6$ , $b = 15$ , $c = -10$ , $d = 10$ , $e = 6$ , $f = 15$ , $g = 19$ , $h = 50$ , $i = -30$ , $j = 20$ , and $a + b + c + d + e + f + g + h + i + j = \boxed{89}$ .

The image $\mathbf{p'}$ of a point $\mathbf{p}$ under a reflection across the plane $\mathbf{n}^T (\mathbf{r} - \mathbf{r}_0) = 0$ is given by

$\mathbf{p'} = \mathbf{r}_0 + \mathbf{M} (\mathbf{p} - \mathbf{r}_0) = \mathbf{M} \mathbf{p} + \mathbf{r}_0 - \mathbf{M} \mathbf{r}_0$

where the reflection matrix $\mathbf{M} = \mathbf{I} - 2 \hspace{4pt} \dfrac{\mathbf{n} \mathbf{n}^T }{ \mathbf{n}^T \mathbf{n} }$

In this problem, the reflection plane is $5x - 3y + 2z = 10$ , therefore, $\mathbf{n} = (5, -3, 2 )$ , and we can take $\mathbf{r}_0 = (0, 0, 5)$

The reflection matrix is, therefore,

$\mathbf{M} = \begin{bmatrix} 1 && 0 && 0\\ 0 && 1 && 0 \\ 0 && 0 && 1 \end{bmatrix} - \dfrac{2}{5^2 + (-3)^2 + 2^2 } \begin{bmatrix} 5 \\ -3 \\ 2 \end{bmatrix} \begin{bmatrix} 5 && -3 && 2 \end{bmatrix}$

Simplifying,

$\mathbf{M} = \begin{bmatrix} 1 && 0 && 0 \\ 0 && 1 && 0 \\ 0 && 0 && 1 \end{bmatrix} - \dfrac{1}{19} \begin{bmatrix} 25 && -15 && 10 \\ -15 && 9 && - 6 \\10 && - 6 && 4 \end{bmatrix}$

Which becomes,

$\mathbf{M} = \dfrac{1}{19} \begin{bmatrix} -6 && 15 && - 10 \\ 15 && 10 && 6 \\ -10 && 6 && 15 \end{bmatrix}$

Vector $\mathbf{v} = \mathbf{r}_0 - \mathbf{M} \mathbf{r}_0$ , that is,

$\mathbf{v} = \begin{bmatrix} 0 \\ 0 \\ 5 \end{bmatrix} - \dfrac{1}{19} \begin{bmatrix} -6 && 15 && - 10 \\ 15 && 10 && 6 \\ -10 && 6 && 15 \end{bmatrix} \begin{bmatrix} 0 \\ 0 \\ 5 \end{bmatrix}$

and this evaluates to,

$\mathbf{v} = \dfrac{1}{19} \begin{bmatrix} 50 \\ -30 \\ 20 \end{bmatrix}$

The required sum is therefore,

$(-6) + 15 + (-10) + 10 + 6 + 15 + 19 + 50 + (-30) + 20 = \boxed{89}$