1. **Find the equation of the line of intersection of the planes:** Given planes: $$x - y + 2z = 3$$ $$x + 2y + z = 2$$ 2. To find the line of intersection, we solve the system of equations simultaneously. 3. Subtract the second equation from the first: $$ (x - y + 2z) - (x + 2y + z) = 3 - 2 $$ $$ x - y + 2z - x - 2y - z = 1 $$ $$ -3y + z = 1 $$ 4. Express $z$ in terms of $y$: $$ z = 1 + 3y $$ 5. Substitute $z$ into the first plane equation: $$ x - y + 2(1 + 3y) = 3 $$ $$ x - y + 2 + 6y = 3 $$ $$ x + 5y = 1 $$ 6. Express $x$ in terms of $y$: $$ x = 1 - 5y $$ 7. Let $y = t$ (parameter), then: $$ x = 1 - 5t $$ $$ y = t $$ $$ z = 1 + 3t $$ 8. The parametric equations of the line of intersection are: $$ x = 1 - 5t, y = t, z = 1 + 3t $$ --- 9. **Find the intersection point of the line $L$:** $$ x = 4 - t, y = 6, z = 7 + 2t $$ with the plane: $$ x - y + z = 1 $$ 10. Substitute the parametric coordinates into the plane equation: $$ (4 - t) - 6 + (7 + 2t) = 1 $$ $$ 4 - t - 6 + 7 + 2t = 1 $$ $$ (4 - 6 + 7) + (-t + 2t) = 1 $$ $$ 5 + t = 1 $$ 11. Solve for $t$: $$ t = 1 - 5 = -4 $$ 12. Find the intersection point by substituting $t = -4$ into the line equations: $$ x = 4 - (-4) = 8 $$ $$ y = 6 $$ $$ z = 7 + 2(-4) = 7 - 8 = -1 $$ 13. Intersection point is: $$ (8, 6, -1) $$ --- 14. **Find the distance between the planes:** $$ x + y + z = 3 $$ $$ 5x + 5y + 5z = 1 $$ 15. Normalize the second plane equation by dividing by 5: $$ x + y + z = \frac{1}{5} $$ 16. The planes are parallel because their normal vectors are proportional. 17. Distance between two parallel planes: $$ d = \frac{|d_1 - d_2|}{\sqrt{a^2 + b^2 + c^2}} $$ where $a, b, c$ are coefficients of $x, y, z$ and $d_1, d_2$ are constants on the right side. 18. Here: $$ a = 1, b = 1, c = 1 $$ $$ d_1 = 3, d_2 = \frac{1}{5} $$ 19. Calculate distance: $$ d = \frac{|3 - \frac{1}{5}|}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{\frac{15}{5} - \frac{1}{5}}{\sqrt{3}} = \frac{\frac{14}{5}}{\sqrt{3}} = \frac{14}{5\sqrt{3}} $$ 20. Rationalize the denominator: $$ d = \frac{14}{5\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{14\sqrt{3}}{15} $$ **Final answers:** - Line of intersection: $$ x = 1 - 5t, y = t, z = 1 + 3t $$ - Intersection point of line and plane: $$ (8, 6, -1) $$ - Distance between planes: $$ \frac{14\sqrt{3}}{15} $$