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## The given polynomial is f(x) = $\left(k-1\right){x}^{2}+kx+1$. Since $-3$ is one of the zeroes of the given polynomial, so $f\left(-3\right)=0$. $\left(k-1\right){\left(-3\right)}^{2}+k\left(-3\right)+1=0\phantom{\rule{0ex}{0ex}}⇒9\left(k-1\right)-3k+1=0\phantom{\rule{0ex}{0ex}}⇒9k-9-3k+1=0\phantom{\rule{0ex}{0ex}}⇒6k-8=0\phantom{\rule{0ex}{0ex}}⇒k=\frac{4}{3}$ Hence, the correct answer is option A.

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