Let the co-ordinates of one vertex of Δ ABC be A(0,2,α ) and the other two vertices lie on the line x+α 5= y-1 2= z+4 3. For α ∈ mathbbZ, if the area…

Given,

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

Now plotting the diagram we get,

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

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

$$\equiv (5k-\alpha ,2k-1,3k-4-\alpha )$$≡(5k−α,2k−1,3k−4−α)

Now $AD$AD is perpendicular to $BC$BC,

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

$$\Rightarrow 19k-4\alpha -7=0.....(1)$$⇒19k−4α−7=0.....(1)

Also given area, $21=\frac{1}{2}\times 2\sqrt{21}\times AD$21=21​×221​×AD

$$\Rightarrow AD=\sqrt{21}$$⇒AD=21​

$$\Rightarrow (5k-\alpha )+(2k-1)+(3k-4-\alpha )=21$$⇒(5k−α)+(2k−1)+(3k−4−α)=21

$$\Rightarrow 19{k}^{2}-8k\alpha +{\alpha }^{2}-14k+4\alpha =2......(2)$$⇒19k2−8kα+α2−14k+4α=2......(2)

Now on solving equation (1)&(2) we get,

$\alpha =3\Rightarrow {\alpha }^{2}=9$α=3⇒α2=9.