3D Coordinate Geometry

Set up one equation per coordinate from $A(2,-4,6)$ and $M(4,0,2)$: $$\frac{2+x_B}{2}=4,\qquad \frac{-4+y_B}{2}=0,\qquad \frac{6+z_B}{2}=2$$

Multiply each equation by $2$: $$2+x_B=8,\qquad -4+y_B=0,\qquad 6+z_B=4$$

Solve each one: $$x_B=6,\qquad y_B=4,\qquad z_B=-2$$

Check: the midpoint of $A(2,-4,6)$ and $(6,4,-2)$ is $\left(\dfrac{2+6}{2},\ \dfrac{-4+4}{2},\ \dfrac{6-2}{2}\right)=(4,0,2)=M$. $$B(6, 4, -2)$$

$d = \sqrt{3^2 + 4^2 + 0^2} = \sqrt{25} = 5$. (3-4-5 triple in the $xy$-plane.)

$1\cdot 2 + 2\cdot(-1) + 3\cdot 1 = 2 - 2 + 3 = 3$.

$\|\vec{v}\| = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.

Average each coordinate: $((2+8)/2, (-1+5)/2, (4-2)/2) = (5, 2, 1)$.

Locate the endpoint of the median, the midpoint $M$ of $BC$ with $B(5,0,-1)$ and $C(-1,4,3)$: $$M=\left(\frac{5+(-1)}{2},\ \frac{0+4}{2},\ \frac{-1+3}{2}\right)=(2, 2, 1)$$

Take the coordinate differences between $A(1,2,3)$ and $M(2,2,1)$: $$\Delta x = 2-1 = 1,\qquad \Delta y = 2-2 = 0,\qquad \Delta z = 1-3 = -2$$

Square, add, and take the square root: $$|AM|=\sqrt{1^2+0^2+(-2)^2}=\sqrt{1+0+4}=\sqrt{5}$$

A general point of $\ell$ has coordinates

$$(x, y, z) = (1 + 2t,\; 2 - t,\; -1 + 3t).$$

Such a point lies on $\pi$ exactly when its coordinates satisfy $x + y + z = 10$:

$$(1 + 2t) + (2 - t) + (-1 + 3t) = 10.$$

Collecting the constants and the terms in $t$:

$$2 + 4t = 10 \quad \Longrightarrow \quad 4t = 8 \quad \Longrightarrow \quad t = 2.$$

Substituting $t = 2$ back into the parametric coordinates:

$$x = 1 + 2(2) = 5, \qquad y = 2 - 2 = 0, \qquad z = -1 + 3(2) = 5.$$

The check $5 + 0 + 5 = 10$ confirms the point lies on $\pi$, and $t = 2$ places it on $\ell$. Since the direction vector $(2, -1, 3)$ is not perpendicular to the normal $(1, 1, 1)$ — their dot product is $2 - 1 + 3 = 4 \neq 0$ — the line is not parallel to the plane, so this intersection point is unique:

$$(5, 0, 5).$$

$(1, 0, 0) \times (0, 2, 0) = (0\cdot 0 - 0\cdot 2, 0\cdot 0 - 1\cdot 0, 1\cdot 2 - 0\cdot 0) = (0, 0, 2)$.

$3a - 1 - 2 = 0 \Rightarrow 3a = 3 \Rightarrow a = 1$.

$d = |{-6}|/\sqrt{4 + 4 + 1} = 6/3 = 2$.