$\therefore \frac{a}{\left|\begin{array}{ll}2 & 3 \\ 2 & 1\end{array}\right|}=\frac{b}{\left|\begin{array}{ll}3 & 1 \\ 1 & 3\end{array}\right|}=\frac{c}{\left|\begin{array}{ll}1 & 2 \\ 3 & 2\end{array}\right|}$
$\therefore \frac{a}{-4}=\frac{b}{8}=\frac{c}{-4}$
i.e. $\frac{a}{1}=\frac{b}{-2}=\frac{c}{1}$
∴ a, b, c are proportional to 1, -2, 1 ∴ from (1), the required cartesian equation is x – 2y + z = 0. Case 2 : Let the plane make non zero intercept p on each axis.
then its equation is $\frac{x}{p}+\frac{y}{p}+\frac{z}{p}=1$
i.e. x + y + z = p …(2) Since this plane pass through (1, 2, 3) and (3, 2, 1) ∴ 1 + 2 + 3 = p and 3 + 2 + 1 = p ∴ p = 6 ∴ from (2), the required cartesian equation is x + y + z = 6 Hence, the cartesian equations of required planes are x + y + z = 6 and x – 2y + z = 0.
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