Let $\frac{x^2 - 6x + 10}{3 - x} = p$
$x^2 - 6x + 10 = 3p - px$
$x^2 - (6 - p)x + 10 - 3p = 0$
Since the equation will have real roots,
$(6 - p)^2 - 4 \times (10 - 3p) \ge 0$
$p^2 - 12p + 12p + 36 - 40 \ge 0$
$p^2 \ge 4$
$p \ge 2 ,\ p \le -2$
Now, when $p = -2$, $x = 4$. Since it is given that $x < 3$, thus this value will be discarded.
Now, $\frac{1}{2}$ and $-\frac{1}{2}$ do not come in the mentioned range.
When $p = 2$, $x = 2$
Thus, the minimum possible value of p will be 2.
Thus, the correct option is C.
Alternate explanation:
Since $x < 3$,
$3 - x > 0$
Let $3 - x = y$. So, $y > 0$.
Now,
$\frac{x^2 - 6x + 10}{3 - x} = \frac{x^2 - 6x + 9 + 1}{3 - x}$
$\Rightarrow \frac{(3 - x)^2 + 1}{3 - x}$
Since $3 - x = y$, the equation will transform to $\frac{y^2 + 1}{y}$ or $y + \frac{1}{y}$
The minimum value of the expression $y + \frac{1}{y}$ for $y > 0$ will be at $y = 1$
i.e. Minimum value = $1 + 1 = 2$
Thus, the correct option is C.
