Step 1: Factor the second pair of lines.
For $2x^2 + bxy - 3y^2 = 0$, this factors because $-3 \times 2 = -6$ and we need factors. Try $(2x - y)(x + 3y) = 2x^2 + 6xy - xy - 3y^2 = 2x^2 + 5xy - 3y^2$. So $b = 5$ gives lines $y = 2x$ and $x + 3y = 0$, i.e., $y = -x/3$. These are NOT perpendicular ($2 \times (-1/3)
e -1$). Try $b = 1$: $(2x - 3y)(x + y)= 2x^2 + 2xy - 3xy - 3y^2 = 2x^2 - xy - 3y^2$. Not matching.
Step 2: Find the common line by factoring the second pair directly.
$2x^2 + bxy - 3y^2 = (2x - y)(x + 3y)$ when $b = 5$ gives slopes $m = 2$ and $m = -1/3$. Check: $m_1 \times m_2 = 2 \times (-1/3) = -2/3
e -1$. Try $(x + y)(2x - 3y) = 2x^2 - 3xy + 2xy - 3y^2 = 2x^2 - xy - 3y^2$ with $b = -1$. Also try $(2x + 3y)(x - y) = 2x^2 - 2xy + 3xy - 3y^2 = 2x^2 + xy - 3y^2$ with $b = 1$: slopes $m = -3/2$ and $m = 1$. Check: $(-3/2)(1) = -3/2
e -1$.
Step 3: The common line must satisfy both pairs.
Let $y = mx$ be the common line. For pair 1: $3m^2 + am + 2 = 0$. For pair 2: $-3m^2 + bm + 2 = 0$. Adding: $(a+b)m + 4 = 0$, so $m = -4/(a+b)$.
Step 4: Use the perpendicularity condition.
For pair 1 ($3m^2 + am + 2 = 0$) with roots $m_1, m_2$: product $m_1 m_2 = 2/3$. If $m_1 = m$ (common), then the other root of pair 1 is $m_3 = 2/(3m)$. For pair 2 ($-3m^2 + bm + 2 = 0$) with roots $m, m_4$: product $m \cdot m_4 = -2/3$, so $m_4 = -2/(3m)$. The condition is $m_3 \cdot m_4 = -1$ (perpendicular): \[ \frac{2}{3m} \cdot \frac{-2}{3m} = -1 \Rightarrow \frac{-4}{9m^2} = -1 \Rightarrow m^2 = \frac{4}{9} \Rightarrow m = \pm \frac{2}{3}. \]
Step 5: Find $a$ and $b$.
Use $m = 2/3$ in $3m^2 + am + 2 = 0$: $3(4/9) + a(2/3) + 2 = 0 \Rightarrow 4/3 + 2a/3 + 2 = 0 \Rightarrow 2a/3 = -10/3 \Rightarrow a = -5$. Wait, let us check $m = -2/3$: $3(4/9) + a(-2/3) + 2 = 0 \Rightarrow 4/3 - 2a/3 + 2 = 0 \Rightarrow -2a/3 = -10/3 \Rightarrow a = 5$.
Step 6: Find $b$ using pair 2.
$-3(4/9) + b(-2/3) + 2 = 0 \Rightarrow -4/3 - 2b/3 + 2 = 0 \Rightarrow 2/3 = 2b/3 \Rightarrow b = 1$. So $a = 5$, $b = 1$.
\[ \boxed{a = 5,\ b = 1} \]