Four posts stand at A(3, 5, 9), B(-5, 4, 6), C(4, -4, 5) and D(-1, -1, -3), with distances in…

Step 1. The three equations. With the formula for two points, using $|A|^2 = 115$, $|B|^2 = 77$, $|C|^2 = 57$ and $|D|^2 = 11$:

• $PA = PB$: $-16x - 2y - 6z = -38$, which is $8x + y + 3z = 19$.
• $PA = PC$: $2x - 18y - 8z = -58$, which is $x - 9y - 4z = -29$.
• $PA = PD$: $-8x - 12y - 24z = -104$, which is $2x + 3y + 6z = 26$.

Step 2. Solve. From the first equation $y = 19 - 8x - 3z$. Put this in the third: $22x + 3z = 31$. Put it in the second: $73x + 23z = 142$. From $22x + 3z = 31$ we get $z = \dfrac{31 - 22x}{3}$, and then $219x + 713 - 506x = 426$, so $x = 1$. Then $z = 3$ and $y = 19 - 8 - 9 = 2$. So $P = (1, 2, 3)$.

Step 3. The distance. $PA^2 = 4 + 9 + 36 = 49$, so the distance is $7\ \text{m}$.