Let $P$ be the plane passing through the points $\left(5,3,0\right),\left(13,3,-2\right)$ and $\left(1,6,2\right)$. For $\alpha \in \mathrm{\mathbb{N}}$, if the distance of the points$A\left(3,4,\alpha \right)$ and $B\left(2,\alpha ,a\right)$ from the plane $P$ are $2$ and $3$ respectively, then the positive value of $a$ is

Given,

$P$ be the plane passing through the points $\left(5,3,0\right),\left(13,3,-2\right)$ and $\left(1,6,2\right)$

Now, $\overrightarrow{AB}=8\hat{i}-2\hat{k}$, $\overrightarrow{AC}=-4\hat{i}+3\hat{j}+2\hat{k}$ ,

So, normal vector to the plane will be,

$\overrightarrow{AB}\times \overrightarrow{AC}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 8& 0& -2\\ -4& 3& 2\end{array}\right|$$=6\hat{i}-8\hat{j}+24\hat{k}$

Now equation of plane will be: $6x-8y+24z=d$ 

And given plane passes through $\left(5,3,0\right)$

So, $6\times 5-8\times 3+24\times 0=d$$\Rightarrow d=6$

Hence, equation of plane will be,
$6x-8y+24z=6\Rightarrow 3x-4y+12z=3$

Now using the formula distance of point $\left(3, 4, \alpha \right)$ from the plane $3x-4y+12z=3$, we get,

$\frac{9-16+12\alpha -3}{\sqrt{9+16+144}}=2\Rightarrow \alpha =3$

Now again using the distance of point $\left(2, \alpha , a\right)\equiv \left(2,3,a\right)$ from the plane formula we get,

$\frac{3\times 2-4\times 3+12\times a-3}{13}=3$

$\Rightarrow 12a-9=39$

$\Rightarrow 12a=48$

$\Rightarrow a=4$