Let the co-ordinates of one vertex of triangle A B C 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 ∈ Z, if the area of triangle A B C is 21 sq. units and the line segment B C has length 2 √21 units, then α2 is equal to

Question

Let the co-ordinates of one vertex of triangle A B C 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 ∈ Z, if the area of triangle A B C is 21 sq. units and the line segment B C has length 2 √21 units, then α2 is equal to (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/f2fa189e808641d0c573f66bc9065bbc-.png" /> |(1/2) ⋅ 2 √21 | i j k α 1 α+4 5 2 3| (1/√25+4+9)|=21 √21 √(2 α+5)2+(2 α+20)2+(2 α-5)2=√21 √38 ⇒ 12 α2+80 α+450=798 ⇒ 12 α2+80 α-348=0 ⇒ α=3 ⇒ α2=9

Q. Let the co-ordinates of one vertex of △ABC be A(0,2,α) and the other two vertices lie on the line 5x+α​=2y−1​=3z+4​, For α∈Z, if the area of △ABC is 21 sq. units and the line segment BC has length 221​ units, then α2 is equal to______

∣∣​21​⋅221​∣∣​iα5​j12​kα+43​∣∣​25+4+9​1​∣∣​=2121​ (2α+5)2+(2α+20)2+(2α−5)2​=21​38​ ⇒12α2+80α+450=798 ⇒12α2+80α−348=0 ⇒α=3⇒α2=9

$∣ ∣ \frac{1}{2} \cdot 221 ∣ ∣ i α 5 j 12 k α + 4 3 ∣ ∣ \frac{1}{25 + 4 + 9} ∣ ∣ = 2121$
$(2 α + 5)^{2} + (2 α + 20)^{2} + (2 α - 5)^{2} = 2138$
$\Rightarrow 12 α^{2} + 80 α + 450 = 798$
$\Rightarrow 12 α^{2} + 80 α - 348 = 0$
$\Rightarrow α = 3 \Rightarrow α^{2} = 9$