Trigonometry!

By Cauchy Schwartz Inequality ,

$(0.6^2 +1^2+ 0.8^2)(\sin^2 x \cos^2 y + \cos^2 x + \sin^2 x \sin^2 y) \geq (0.6\sin x \cos y + \cos x + 0.8\sin x \sin y )^2$

$\implies 2 \geq 2$

Thus since equality holds, we get $\sin^2x =1$ and $\sin^2 y = 0.64$ making final answer to be $150.$

Consider the vectors $\overrightarrow{ \rm A} = \sin{x} \cos{y}~ \hat{i} + \cos{x} ~ \hat{j} + \sin{x} \sin{y}~ \hat{k}$ and $\overrightarrow{ \rm B } = 0.6 ~\hat{i} + 1~\hat{j }+ 0.8~ \hat{ k}$

Angle between the vectors would be $cos{ \theta} = \dfrac{ \overrightarrow{\rm A} \cdot \overrightarrow{ \rm B } }{ \mod{ \overrightarrow{\rm A} } × \mod{ \overrightarrow { \rm B } } }$

Note that $\mod{ \overrightarrow{ \rm A} }= 1 , \mod{ \overrightarrow{ \rm B }} = \sqrt{2}$ . Thus we get

$\cos{ \theta} = \dfrac{ 0.6 \sin{x} \cos{y} + \cos{x} + 0.8 \sin{x} \sin{y}}{ \sqrt{2} }$

From the given equation, we get that $cos { \theta} = 1$ .Thus the lines are $\color{#3D99F6}{parallel}$ !!!

$\therefore \dfrac { \sin{x} \cos{y} }{ 0.6} = \dfrac{ \cos{x} }{ 1} = \dfrac{ \sin{x} \sin{y}}{ 0.8}$

On solving, we get $\tan^2{x} = 1 , \tan^2{y} = \dfrac{16}{9}$

Thus $\color{#3D99F6}{ 6 \tan^2{x} + 81\tan^2{y} } = 150$