The problem asks to find the value of $n$ in triangle $PQR$, where the measures of the angles are given as $m\angle P = 4n + 61$, $m\angle Q = 67 - 3n$, and $m\angle R = n + 74$. Also, we need to list the sides of the triangle from shortest to longest.

1. Problem Description

The problem asks to find the value of $n$ in triangle $PQR$, where the measures of the angles are given as $m\angle P = 4n + 61$, $m\angle Q = 67 - 3n$, and $m\angle R = n + 74$. Also, we need to list the sides of the triangle from shortest to longest.

2. Solution Steps

First, we need to find the value of $n$. We know that the sum of the angles in a triangle is 180 degrees. Therefore, we can write the equation:

$m\angle P + m\angle Q + m\angle R = 180$

Substitute the given expressions for the angles:

$(4n + 61) + (67 - 3n) + (n + 74) = 180$

Combine like terms:

$(4n - 3n + n) + (61 + 67 + 74) = 180$

$2n + 202 = 180$

Subtract 202 from both sides:

$2n = 180 - 202$

$2n = -22$

Divide by 2:

$n = -11$

Now, we need to find the measures of each angle:

$m\angle P = 4(-11) + 61 = -44 + 61 = 17$

$m\angle Q = 67 - 3(-11) = 67 + 33 = 100$

$m\angle R = (-11) + 74 = 63$

We have the angles: $m\angle P = 17^{\circ}$, $m\angle Q = 100^{\circ}$, and $m\angle R = 63^{\circ}$.

In a triangle, the shortest side is opposite the smallest angle, and the longest side is opposite the largest angle. So, since $m\angle P < m\angle R < m\angle Q$, we have:

$17^{\circ} < 63^{\circ} < 100^{\circ}$

Therefore, $\angle P < \angle R < \angle Q$. The sides opposite these angles are $QR$, $PQ$, and $PR$, respectively. So, the order of the sides from shortest to longest is: $QR$, $PQ$, $PR$.

3. Final Answer

$n = -11$

Sides from shortest to longest: $QR$, $PQ$, $PR$