Explanation

(a)
On solving line Ist and IInd, and Ist and IIIrd, we get $A( - 3,4)\$and $B\left( - \frac{3}{5},\frac{8}{5} \right)$.
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The equation of perpendicular line to the line $4x - y + 4 = 0$ and passes through the point $A( - 3,\ 4)$ is
$x + 4y - 13 = 0$ …(i)
Also, the equation of perpendicular line to the line $2x + 3y = 6$ and passes through $a\ point\ B\left( - \frac{3}{5},\frac{8}{5} \right)\ is\$
$3x - 2y + 5 = 0$ …(ii)
On solving Eq. (i) and (ii), we get the orthocentre $\left( \frac{3}{7},\frac{22}{7} \right)$
Which is lies in Ist quadrant.