$\therefore \frac{{\partial u}}{{\partial x}} = \frac{1}{{{x^2} + {y^2}}}.2x$
$\frac{{{\partial ^2}u}}{{\partial {x^2}}} = \frac{{({x^2} + {y^2}).2 - 2x.2x}}{{{{({x^2} + {y^2})}^2}}}$$ = \frac{{2({y^2} - {x^2})}}{{{{({x^2} + {y^2})}^2}}}$’
$\frac{{\partial u}}{{\partial y}} = \frac{1}{{{x^2} + {y^2}}}.2y$
$\therefore \frac{{{\partial ^2}u}}{{\partial {y^2}}} = \frac{{({x^2} + {y^2})\,.\,2 - 2y.2y}}{{{{({x^2} + {y^2})}^2}}} = \frac{{2({x^2} - {y^2})}}{{{{({x^2} + {y^2})}^2}}}$
$\therefore $ $\frac{{{\partial ^2}u}}{{\partial {x^2}}} + \frac{{{\partial ^2}u}}{{\partial {y^2}}} = 0$.
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$a x+2 a y-3 a z=1$
$(2 a+1) x+(2 a+3) y+(a+1) z=2$
$(3 a+5) x+(a+5) y+(a+2) z=3$
has unique solution and infinitely many solutions. Then
$\overline{A B}=-2 \hat{i}+\hat{j}+3 \hat{k}$
$\overline{C B}=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$
$\overline{C A}=4 \hat{i}+3 \hat{j}+\delta \hat{k}$
If $\delta > 0$ and the area of the triangle $ABC$ is $5 \sqrt{6}$, then $\overline{C B} \cdot \overline{C A}$ is equal to