The plane 2 x-y+z=4 intersects the line segment joining the points...

Given points are:

$A ( a, - 2, 4 )$ and $B ( 2, b, - 3 )$

Point $C$ divides the line segment joining points $A$ and $B$ in the ratio $2: 1$, so

$C ≡ \left(\dfrac{a + 4}{3}, \dfrac{2b - 2}{3}, - \dfrac{2}{3}\right)$

Also, $C$ lies on $2x - y + 2 = 4$, therefore

$\dfrac{2a + 8}{3} - \left(\dfrac{2b - 2}{3}\right) - \dfrac{2}{3} = 4$

$⇒ 2a + 8 - 2b + 2 - 2 = 12$

$⇒ a - b = 2 . . . . \left(1\right)$

Also, $OC = \sqrt{5}$

$⇒ {\left(\dfrac{a + 4}{3}\right)}^{2} + {\left(\dfrac{2b - 2}{3}\right)}$^{2} + \dfrac{4}{9} = 5$</p><p>$⇒ {a}^{2} + 8a + 16 + 4{b}^{2} + 4 - 8b = 45$</p><p>$⇒ {a}^{2} + 8a + 4{b}^{2} - 8b = 25 . . . \left(2\right)$</p><p>Solving \left(1\right)$ and $( 2 )$, we have

$( b + 6{)}^{2} + ( 2b - 2{)}^{2} = 41$

$⇒ 5{b}^{2} + 4b - 1 = 0$

$⇒ b = - 1$ or $\dfrac{1}{5}$

$⇒ a = 1$ or $\dfrac{11}{5}$

But $ab < 0 ⇒ \left(a, b\right) ≡ \left(1, - 1\right)$

$C ≡ \left(\dfrac{5}{3}, \dfrac{- 4}{3}, \dfrac{- 2}{3}\right), P ≡ ( 2, - 1, - 3 )$

$C{P}^{2} = \dfrac{1}{9} + \dfrac{1}{9} + \dfrac{49}{9} = \dfrac{51}{9} = \dfrac{17}{3}$