Given,

The co-ordinates of one vertex of $∆ABC$ be $A(0,2,\alpha )$ and the other two vertices lie on the line $\frac{\mathrm{x}+\alpha }{5}=\frac{\mathrm{y}-1}{2}=\frac{\mathrm{z}+4}{3}$

Now plotting the diagram we get,

Now any point on the plane $\frac{\mathrm{x}+\alpha }{5}=\frac{\mathrm{y}-1}{2}=\frac{\mathrm{z}+4}{3}=k$ is given by $\left(5k-\alpha , 2k+1, 3k-4\right)$

Now direction ratio of line $AD$ is given by, $\left(5k-\alpha -0, 2k+1-2, 3k-4-\alpha \right)$

$\equiv \left(5k-\alpha , 2k-1, 3k-4-\alpha \right)$

Now $AD$ is perpendicular to $BC$,

So, $5\left(5k-\alpha \right)+2\left(2k-1\right)+3\left(3k-4-\alpha \right)=0$

$\Rightarrow 19k-4\alpha -7=0 .....\left(1\right)$

Also given area, $21=\frac{1}{2}\times 2\sqrt{21}\times AD$

$\Rightarrow AD=\sqrt{21}$

$\Rightarrow \left(5k-\alpha \right)+\left(2k-1\right)+\left(3k-4-\alpha \right)=21$

$\Rightarrow 19k^2-8k\alpha +{\alpha }^2-14k+4\alpha =2 ......\left(2\right)$

Now on solving equation $\left(1\right) \& \left(2\right)$ we get,

$\alpha =3\Rightarrow {\alpha }^2=9$.