Correct option is (D) 13
Let \(7\alpha\;and\;6\alpha\) are roots of quadratic equation \(3x^2-kx + 14= 0.\)
\(\therefore\) Product of roots \(=\frac{14}3\)
\(\Rightarrow\) \(7\alpha\times6\alpha\) \(=\frac{14}3\)
\(\Rightarrow\) \(\alpha^2=\frac{14}{3\times42}\)
\(=\frac1{3\times3}=\frac19\)
\(\Rightarrow\) \(\alpha=\pm\frac13\) _______________(1)
Now, sum of roots \(=\frac{-(-k)}3=\frac{k}3\)
\(\Rightarrow\) \(7\alpha+6\alpha\) \(=\frac k3\)
\(\Rightarrow\) \(13\alpha\) \(=\frac k3\)
\(\Rightarrow\) \(\frac k3=\pm\frac{13}3\) (From (1))
\(\Rightarrow k=\pm13\)
\(\Rightarrow\) k = 13 or k = -13
