Step 1: Equation of the plane with intercepts.
- The general equation of a plane with intercepts \(a\), \(b\), and \(c\) on the \(X\), \(Y\), and \(Z\)-axes, respectively, is:
\[
\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1
\]
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Step 2: Substitute the point \((1, 0, -2)\) into the plane equation.
- Since the point \((1, 0, -2)\) lies on the plane, substitute \(x = 1\), \(y = 0\), and \(z = -2\) into the equation:
\[
\frac{1}{a} + \frac{0}{b} + \frac{-2}{c} = 1
\]
\[
\frac{1}{a} - \frac{2}{c} = 1
\]
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Step 3: Find the normal vector of the plane.
- The foot of the perpendicular from the point \((-2, -1, 3)\) to the plane is \((1, 0, -2)\), so the vector from \((-2, -1, 3)\) to \((1, 0, -2)\) is:
\[
\vec{v} = (1 - (-2), 0 - (-1), -2 - 3) = (3, 1, -5)
\]
- This vector \(\vec{v} = (3, 1, -5)\) is parallel to the normal vector of the plane. Thus, the normal vector of the plane is \(\vec{n} = (3, 1, -5)\).
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Step 4: Use the point-normal form of the plane equation.
- The point-normal form of the equation of the plane is:
\[
3(x - 1) + 1(y - 0) - 5(z + 2) = 0
\]
Simplifying this:
\[
3(x - 1) + y - 5(z + 2) = 0
\]
\[
3x - 3 + y - 5z - 10 = 0
\]
\[
3x + y - 5z - 13 = 0
\]
So, the equation of the plane is:
\[
3x + y - 5z = 13
\]
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Step 5: Identify the intercepts of the plane.
- The intercepts of the plane are the points where the plane intersects the axes:
- Set \(y = 0\) and \(z = 0\) to find the \(x\)-intercept:
\[
3x = 13 \quad \Rightarrow x = \frac{13}{3}
\]
- Set \(x = 0\) and \(z = 0\) to find the \(y\)-intercept:
\[
y = 13
\]
- Set \(x = 0\) and \(y = 0\) to find the \(z\)-intercept:
\[
-5z = 13 \quad \Rightarrow z = -\frac{13}{5}
\]
Thus, the intercepts are:
\[
a = \frac{13}{3}, \quad b = 13, \quad c = -\frac{13}{5}
\]
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Step 6: Calculate \(3a + b + 5c\).
- Finally, calculate:
\[
3a + b + 5c = 3 \times \frac{13}{3} + 13 + 5 \times \left(-\frac{13}{5}\right)
\]
\[
= 13 + 13 - 13 = 13
\]
Thus, \(3a + b + 5c = 13\).
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