Let's recall that, the potential difference between any two points X(x,y,z) and Y(a,b,c) is given by ;
- [tex]{\boxed{\bf{V_{Y}-V_{X}=\displaystyle \bf -\int_{X}^{Y}\overrightarrow{E}\cdot \overrightarrow{dr}}}}[/tex]
So, here ;
[tex]{:\implies \quad \sf \overrightarrow{E}=3x\hat{i}-2y\hat{j}+5z\hat{k}}[/tex]
So, now our second component of the Integrand will just be ;
[tex]{:\implies \quad \sf \overrightarrow{dr}=dx\hat{i}+dy\hat{j}+dz\hat{k}}[/tex]
So, now the whole integrand will just be ;
[tex]{:\implies \quad \sf \overrightarrow{E}\cdot \overrightarrow{dr}=(3x\hat{i}-2y\hat{j}+5z\hat{k})(dx\hat{i}+dy\hat{j}+dz\hat{k})}[/tex]
[tex]{:\implies \quad \sf \overrightarrow{E}\cdot \overrightarrow{dr}=3xdx-2ydy+5zdz}[/tex]
Now, Let's move to the final answer ;
[tex]{:\implies \quad \displaystyle \sf V_{Y}-V_{X}=-\int_{X}^{Y}3xdx-2ydy+5zdz}[/tex]
As,X is the point (1,3,5) and Y being (3,2,7) , so seperate the integral into three integrals with limits as follows respectively;
[tex]{:\implies \quad \displaystyle \sf V_{Y}-V_{X}=-\bigg(\int_{1}^{3}3xdx-\int_{3}^{2}ydy+\int_{5}^{7}zdz\bigg)}[/tex]
[tex]{:\implies \quad \displaystyle \sf V_{Y}-V_{X}=-\bigg(3\int_{1}^{3}xdx-2\int_{3}^{2}ydy+5\int_{5}^{7}zdz\bigg)}[/tex]
[tex]{:\implies \quad \displaystyle \sf V_{Y}-V_{X}=-\bigg\{3\bigg(\dfrac{x^2}{2}\bigg)\bigg|_{1}^{3}-2\bigg(\dfrac{y^2}{2}\bigg)\bigg|_{3}^{2}+5\bigg(\dfrac{z^2}{2}\bigg)\bigg|_{5}^{7}\bigg\}}[/tex]
[tex]{:\implies \quad \displaystyle \sf V_{Y}-V_{X}=-\bigg\{2\bigg(\dfrac{9}{2}-\dfrac12\bigg)-2\bigg(\dfrac{4}{2}-\dfrac92\bigg)+5\bigg(\dfrac{49}{2}-\dfrac{25}{2}\bigg)\bigg\}}[/tex]
[tex]{:\implies \quad \displaystyle \sf V_{Y}-V_{X}=-\{3(4)-(-5)+5(12)\}}[/tex]
[tex]{:\implies \quad \displaystyle \sf V_{Y}-V_{X}=-\{12+5+60\}}[/tex]
[tex]{:\implies \quad \displaystyle \boxed{\bf{V_{Y}-V_{X}=-77\:\: Volt}}}[/tex]
Hence, this is the required answer
