The coordinate of the point(s) on the graph of the function, $f\left(x\right)=\frac{x^3}{3}-\frac{5x^2}{2}7x-4$ where the tangent drawn cuts-off intercepts from the coordinate axes which are equal in magnitude but opposite in sign, is?

 Given $f\left(x\right)=\frac{x^3}{3}-\frac{5x^2}{2}+7x-4.$ Let the required point be $P.$

Since, intercepts are equal in magnitude but opposite in sign,

${\left(\frac{dy}{dx}\right)}_P=1$

Now, By differentiating $f\left(x\right)$ w.r.t x we get: 

$$\begin{gathered} \frac{dy}{dx}=x^2-5x+7 \\ \Rightarrow x^2-5x+7=1 \end{gathered}$$

$$\begin{gathered} \Rightarrow x^2-5x+6=0 \\ \Rightarrow x=2 or 3 \end{gathered}$$

If $x=2$ we get:

$$\begin{gathered} y=f\left(2\right)=\frac{2^3}{3}-\frac{5{\left(2\right)}^2}{2}+7\left(2\right)-4 \\ \Rightarrow y=\frac{8}{3}-10+14-4 \\ \Rightarrow y=\frac{8}{3} \end{gathered}$$

So, The required point $P=\left(2,\frac{8}{3}\right)$.

If $x=3$ we get:

$$\begin{gathered} y=f\left(3\right)=\frac{3^3}{3}-\frac{5{\left(3\right)}^2}{2}+7\left(3\right)-4 \\ \Rightarrow y=9+21-4-\frac{45}{2} \\ \Rightarrow y=\frac{7}{2} \end{gathered}$$

So, the required point $P=\left(3,\frac{7}{2}\right)$

Hence the required point at which the tangent makes an equal length of intercept on the coordinate axes is $\left(2,\frac{8}{3}\right)$.

Thus option $1$ is the correct answer.