If m:n is the ratio in which the point left(f | vedclass

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(A) Let the points be $A(2, p, 2)$ and $B(p, -2, p)$. The point $P\left(\frac{8}{5}, -\frac{1}{5}, \frac{8}{5}\right)$ divides $AB$ in the ratio $m:n$

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$p$

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(A) Let the points be $A(2, p, 2)$ and $B(p, -2, p)$. The point $P\left(\frac{8}{5}, -\frac{1}{5}, \frac{8}{5}\right)$ divides $AB$ in the ratio $m:n$. Using the section formula, the $y$-coordinate of $P$ is given by: $\frac{m(-2) + n(p)}{m+n} = -\frac{1}{5}$ $-10m + 5np = -m - n$ $9m = n(5p + 1) \implies \frac{m}{n} = \frac{5p+1}{9}$ Using the $x$-coordinate of $P$: $\frac{m(p) + n(2)}{m+n} = \frac{8}{5}$ $5mp + 10n = 8m + 8n$ $m(5p - 8) = -2n \implies \frac{m}{n} = \frac{-2}{5p-8} = \frac{2}{8-5p}$ Equating the two ratios: $\frac{5p+1}{9} = \frac{2}{8-5p}$ $(5p+1)(8-5p) = 18$ $40p - 25p^2 + 8 - 5p = 18$ $25p^2 - 35p + 10 = 0$ $5p^2 - 7p + 2 = 0$ $(5p-2)(p-1) = 0$ Since $p$ is an integer, $p = 1$. Then $\frac{m}{n} = \frac{5(1)+1}{9} = \frac{6}{9} = \frac{2}{3}$. We need to find $\frac{3m+n}{3n} = \frac{3(m/n) + 1}{3} = \frac{3(2/3) + 1}{3} = \frac{2+1}{3} = 1$. Since $p=1$, the result is $p$.

If $m:n$ is the ratio in which the point $\left(\frac{8}{5}, -\frac{1}{5}, \frac{8}{5}\right)$ divides the line segment joining the points $(2, p, 2)$ and $(p, -2, p)$, where $p$ is an integer, then $\frac{3m+n}{3n} =$

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