The foot of perpendicular from the origin O to a plane P which...

Plotting the diagram of the given data we get,

As $( 2, a, 4 )$ is the foot of perpendicular from origin,

So, direction ratio of normal vector will be $( 2 - 0, a - 0, 4 - 0 ) ≡ ( 2, a, 4 )$

So, equation of plane $P$ will be,

$2 ( x - 2 ) + a ( y - a ) + 4 ( z - 4 ) = 0$

$⇒ 2x + ay + 4z = 20 + {a}^{2}$

$⇒ \dfrac{x}{\dfrac{20 + {a}^{2}}{2}} + \dfrac{y}{\dfrac{20 + {a}^{2}}{a}} + \dfrac{z}{\dfrac{20 + {a}^{2}}{4}} = 1$

Now Volume of tetrahedron $OABC$ is given by,

$V = \dfrac{1}{6} \cdot OA \cdot OB \cdot OC$

$⇒ \dfrac{1}{6}\left(\dfrac{20 + {a}^{2}}{2}\right)\left(\dfrac{20 + {a}^{2}}{a}\right)\left(\dfrac{20 + {a}^{2}}{4}\right) = 144$

$⇒ {\left(\dfrac{20 + {a}^{2}}{12}\right)}^{3} = 4a$

Let $a = 2{k}^{3}$

$⇒ {\left(\dfrac{20 + 4{k}^{6}}{12}\right)}^{3} = 8{k}^{3}$

$⇒ \left(\dfrac{20 + 4{k}^{6}}{12}\right) = 2k$

$⇒ {k}^{6} + 5 = 6k$

$⇒ k = 1 ⇒ a = 2$

So, equation of plane $2x + 2y + 4z = 24$

$⇒ x + y + 2z = 12$

Clearly all three points $( 0, 4, 4 ) , ( 0, 6, 3 ) , ( 2, 2, 4 )$ satisfy the equation of plane but $( 3, 0, 4 )$ does not satisfy plane