MCQ
If $a \ne p,b \ne q,c \ne r$ and $\left| {\,\begin{array}{*{20}{c}}p&b&c\\{p + a}&{q + b}&{2c}\\a&b&r\end{array}\,} \right|$ =$ 0$, then $\frac{p}{{p - a}} + \frac{q}{{q - b}} + \frac{r}{{r - c}} = $
- A$3$
- ✓$2$
- C$1$
- D$0$
= $\left| {\,\begin{array}{*{20}{c}}{p\,\,\,}&{b\,\,}&c\\{a\,\,\,}&{q\,\,}&c\\{a\,\,\,}&{b\,\,}&r\end{array}\,} \right|\, = 0$
Applying ${R_2} \to {R_2} - {R_1}$ and ${R_3} \to {R_3} - {R_1}$
$\left| {\,\begin{array}{*{20}{c}}p&b&c\\{a - p}&{q - b}&0\\{a - p}&0&{r - c}\end{array}\,} \right|\, = 0$
On expansion we get,
$p\,(q - b)(r - c) - b(a - p)(r - c) - c(q - b)(a - p) = 0$
==> $(p - a)(q - b)(r - c)$$\left[ {\frac{p}{{(p - a)}} + \frac{b}{{(q - b)}} + \frac{c}{{(r - c)}}} \right] = 0$
==> $(p - a)(q - b)(r - c)$$\left[ {\frac{p}{{(p - a)}} + \frac{q}{{(q - b)}} - 1 + \frac{r}{{(r - c)}} - 1} \right] = 0$
$\therefore $ $p \ne a,\,q \ne b,\,r \ne c$
$\therefore $ $\frac{p}{{p - a}} + \frac{q}{{q - b}} + \frac{r}{{r - c}} = 2$.
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Statement $-1:$ The substitution $z = y^2$ transforms the above equation into a first order homogenous differential equation.
Statement $-2:$ The solution of this differential equation is ${y^2}{e^{ - {y^2}/x}} = C$.
$12x + by + cz = 0$ ; $ax + 24y + cz = 0$ ; $ax + by + 36z = 0$ .
(where $a$ , $b$ , $c$ are real numbers, $a \ne 12$ , $b \ne 24$ , $c \ne 36$ ).
If system of equation has solution and $z \ne 0$, then value of $\frac{1}{{a - 12}} + \frac{2}{{b - 24}} + \frac{3}{{c - 36}}$ is
(where $C$ is constant of integration)