Statement (A): If the angle between the lines $2x+3y+5=0$  and $3x–ky+10=0$  is $\frac{\pi }{2}$ , then the value of $‘k’$  is 2.

Statement (B): If the angle between the lines $5x+7y–10=0$ and $ax–14y+20=0$ is $0°$, then the value of $a$ is$‘–2’$

Statement (A): $2x+3y+5=0$ ,    $3x–ky+10=0$ 

$a_{1}x+b_{1}y+c_1=0$ ,    $a_{2}x+b_{2}y+c_2=0$

$\cos \text{θ}=\left|\frac{a_1{a}_2+b_1{b}_2}{\sqrt{a_1{ }^2+b_1{ }^2}\sqrt{a_2{ }^2+b_2{ }^2}}\right|$

$\cos 90°=\left|\frac{2\times 3+3\left(-k\right)}{\sqrt{2^2+3^2}\sqrt{3^2+{\left(-k\right)}^2}}\right|$

$0=\left|\frac{6-3k}{\sqrt{13}\sqrt{9+k^2}}\right|$

$6-3k=0$

$k=2$

Statement (B): $5x+7y–10=0$ ,   $ax–14y+20=0$

If the angle between two lines is $0°$

Then the lines are parallel

$5x+7y–10=0$ ,   $ax-14y+20=0$

$a_{1}x+b_{1}y+c_1=0$ ,  $a_{2}x+b_{2}y+c_2=0$

$\frac{a_1}{a_2}=\frac{b_1}{b_2}$

$\frac{5}{7}=\frac{a}{-14}$

$a=\frac{-70}{7}$

$a=-10$