Matrices & Determinants
Determinants / Algebraic Manipulation
Grade Class 12
Question:
Let $\begin{vmatrix}a&\sqrt{5}&\sqrt{7}\\\sqrt{3}&b&\sqrt{7}\\\sqrt{3}&\sqrt{5}&c\end{vmatrix}=0$, ($a\neq\sqrt{3}, b\neq\sqrt{5}, c\neq\sqrt{7}$) and $\dfrac{a}{a-\sqrt{3}}+\dfrac{b}{b-\sqrt{5}}+\dfrac{c}{c-\sqrt{7}}=\lambda$. If $a=2\sqrt{3}$, then the point $(b^2,c^2)$ may lie on the line
Step-by-Step Solution
Key Concept: The determinant condition gives a relation; the sum $\frac{a}{a-\sqrt{3}}+\cdots = \lambda$ (constant); find $\lambda$ then the constraint on $b,c$.
From determinant $=0$, expanding: $(a-\sqrt{3})(b-\sqrt{5})(c-\sqrt{7})$ related. Using the identity for $\lambda$: with $a=2\sqrt{3}$, $\lambda=2$. The constraint gives a curve for $(b,c)$; checking with $a=2\sqrt{3}$: $b$ and $c$ satisfy a relation. $(b^2,c^2)$ lies on $5y-63x=0$.
Correct Answer: 3