The equation of tangent and normal to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at a point $\left(x_1, y_1\right)$ are respectively $\frac{x{x}_1}{a^2}+\frac{y{y}_1}{b^2}=1$ and $\frac{a^{2}x}{x_1}-\frac{b^{2}y}{y_1}=a^2-b^2.$

The given ellipse is $3x^2+5y^2=32, \Rightarrow \frac{3x^2}{32}+\frac{5y^2}{32}=1.$

Hence, the tangent and normal to the ellipse $\frac{3x^2}{32}+\frac{5y^2}{32}=1$ at $\left(2, 2\right)$ are respectively $\frac{3\times 2\times x}{32}+\frac{5\times 2\times y}{32}=1$ and $\frac{32\times x}{3\times 2}-\frac{32\times y}{5\times 2}=\frac{32}{3}-\frac{32}{5}$

Hence, the tangent is $3x+5y=16 ...\left(1\right)$

And, the normal is $5x-3y=4 ...\left(2\right)$

The point where the tangent meets the $x$-axis can be obtained by putting $y=0, \Rightarrow 3x+0=16 \Rightarrow x=\frac{16}{3},$ thus the point $Q\equiv \left(\frac{16}{3}, 0\right).$

Similarly, the point where the normal meets the $x$-axis, is $x=\frac{4}{5},$ thus the point $R\equiv \left(\frac{4}{3}, 0\right).$ 

Tangent and normal intersect at $P\left(2, 2\right),$ and the length of perpendicular from any point on $x$-axis is the absolute value of its $y$-co-ordinate, hence the height of the triangle is $2$ units.

And, hence the required area is $=\frac{1}{2}\left(QR\right)\times h$

$=\frac{1}{2}\times \left(\frac{16}{3}-\frac{4}{5}\right)\times 2=\frac{68}{15}$ sq units.