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The side of the base \$ABC\$ of a regular triangular prism \$ABC A_1B_1C_1\$ is 6, and the height is \$1 \frac{3}{\sqrt{7}}\$. Points \$P\$, \$Q\$, and \$K\$ are located on the edges \$AC\$, \$A_1C_1\$, and \$BB_1\$ respectively. Find the area of the cross-section and the angle between the plane of the base of the prism and the plane of the cross-section.

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Let the lines $FP$ and $ \$

![](https://cdn.mathpix.com/cropped/2024_05_06_6ba7643fcbc3b9c222e2g-50.jpg?height=48&width=1987&top_left_y=113&top_left_x=1)
pentagon $PLKNF$ is the desired section.

![](https://cdn.mathpix.com/cropped/2024_05_06_6ba7643fcbc3b9c222e2g-50.jpg?height=1009&width=2001&top_left_y=227&top_left_x=14)

From the similarity of triangles $MAP$ and $MA_1F$, we find that $ 1 \frac{MA}{MA_1} = \frac{AP}{A_1F} = \frac{1}{3} $, so $ MA = \frac{1}{2} AA_1 = \frac{1}{2} \cdot \frac{3}{\sqrt{7}} = \frac{3}{2\sqrt{7}}, MA_1 = \frac{3}{2} AA_1 = \frac{3}{2} \cdot \frac{3}{\sqrt{7}} = \frac{9}{2\sqrt{7}}. $ Since $K$ is the midpoint of $BB_1$, $BK = KB_1 = AM$.

Let $H$ be the foot of the perpendicular dropped from point $A_1$ to the line $EEF$. Then, by the theorem of three perpendiculars, $MH \perp FE$, so $MFA_1$ is the linear angle of the dihedral angle between the cutting plane

![](https://cdn.mathpix.com/cropped/2024_05_06_6ba7643fcbc3b9c222e2g-50.jpg?height=48&width=1976&top_left_y=1554&top_left_x=1)
$| \cdot A_1E| \cdot |\cos 60^\circ| = \sqrt{81 + 9 - 9 \cdot 3} = 3\sqrt{7}, $ so $ S_{\Delta A_1FE} = \frac{1}{2} EF \cdot A_1H = \frac{1}{2} \cdot 3\sqrt{7} \cdot A_1H = \frac{3\sqrt{7}}{2} A_1H. $ On the other hand, $ S_{\Delta A_1FE} = \frac{1}{2} A_1F \cdot A_1E \cdot \sin 60^\circ = \frac{1}{2} \cdot 3 \cdot 9 \cdot \frac{\sqrt{3}}{2} = \frac{27\sqrt{3}}{4}. $ From the equation $\frac{3\sqrt{7}}{2} A_1H = \frac{27\sqrt{3}}{4}$, we find that $A_1H = \frac{9\sqrt{3}}{2\sqrt{7}}$. Therefore, $\operatorname{tg} \angle MFA_1 = \frac{MA_1}{A_1H} = \frac{9/(2\sqrt{7})}{9\sqrt{3}/(2\sqrt{7})} = \frac{1}{\sqrt{3}}, \angle MFA_1 = 30^\circ. $

![](https://cdn.mathpix.com/cropped/2024_05_06_6ba7643fcbc3b9c222e2g-51.jpg?height=1134&width=829&top_left_y=-1&top_left_x=617)

Through point $B_1$, draw a line parallel to $A_1C_1$ (Fig.2). Let this line intersect the line $B_1C_1$ at point $Q$. From the similarity of triangles $EB_1Q$ and $ \$ \frac{EB_1}{A_1F} = 3 \cdot \frac{3}{9} = 1, $ and from the similarity of triangles $NQB_1$ and $NFC_1$, $\frac{B_1N}{NC_1} = \frac{B_1Q}{FC_1} = \frac{1}{3}. $

Let $P_1$ and $ \$

![](https://cdn.mathpix.com/cropped/2024_05_06_6ba7643fcbc3b9c222e2g-51.jpg?height=57&width=1946&top_left_y=1389&top_left_x=-1)
$PLKNF$ onto the plane $A_1B_1C_1$, so $ S_{P_1L_1B_1NF} = S_{\Delta A_1B_1C_1} - S_{\Delta A_1P_1L_1} - S_{\Delta C_1FN} = S = S_{\Delta A_1B_1C_1} - \frac{AP_1 1 \{

![](https://cdn.mathpix.com/cropped/2024_05_06_6ba7643fcbc3b9c222e2g-51.jpg?height=65&width=1989&top_left_y=1531&top_left_x=0)
$= S_{\Delta A_1B_1C_1} - \frac{1}{6} \cdot \frac{1}{2} S_{\Delta A_1B_1C_1} - \frac{1}{2} \cdot \frac{3}{4} S_{\Delta A_1B_1C_1} = \frac{13}{24} S_{\Delta A_1B_1C_1} = \frac{13}{24} \cdot \frac{A_1B_1^2 \sqrt{3}}{4} = \frac{13}{24} \cdot \frac{36\sqrt{3}}{4} = \frac{39\sqrt{3}}{8}. $ Therefore, $ S_{PLBNF} = \frac{S_{P_1L_1B_1NF}}{\cos \angle MFA_1} = \frac{39\sqrt{3}/8}{\sqrt{3}/2} = \frac{39}{4}. $

## Answer

$ \frac{39}{4} $ $ \frac{\pi}{6} $.