Question

If \( a, b, c \) are the intercepts made on X, Y, Z-axes respectively by the plane passing through the points \( (1, 0, -2) \), \( (3, -1, 2) \), and \( (0, -3, 4) \), then \( 3a + 4b + 7c = \)?

• A. \( -5 \)
• B. \( 5 \)
• C. \( -15 \)
• D. \( 15 \) \bigskip

Hint:

When solving problems with intercepts and planes, it is useful to set up a system of equations and solve step-by-step. Simplify the equations at each stage to avoid errors.

Answer 1

The Correct Option is C

We are given three points: \( (1, 0, -2) \), \( (3, -1, 2) \), and \( (0, -3, 4) \), which lie on the plane. The equation of the plane is of the form: \[ \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \] where \( a \), \( b \), and \( c \) are the intercepts on the X, Y, and Z axes, respectively. Step 1: Set up the system of equations Substitute the coordinates of the given points into the equation of the plane: 1. For the point \( (1, 0, -2) \): \[ \frac{1}{a} + \frac{0}{b} + \frac{-2}{c} = 1 \implies \frac{1}{a} - \frac{2}{c} = 1 \] 2. For the point \( (3, -1, 2) \): \[ \frac{3}{a} + \frac{-1}{b} + \frac{2}{c} = 1 \implies \frac{3}{a} - \frac{1}{b} + \frac{2}{c} = 1 \] 3. For the point \( (0, -3, 4) \): \[ \frac{0}{a} + \frac{-3}{b} + \frac{4}{c} = 1 \implies - \frac{3}{b} + \frac{4}{c} = 1 \] Thus, we have the following system of equations: \[ \frac{1}{a} - \frac{2}{c} = 1 \quad \text{(Equation 1)} \] \[ \frac{3}{a} - \frac{1}{b} + \frac{2}{c} = 1 \quad \text{(Equation 2)} \] \[ - \frac{3}{b} + \frac{4}{c} = 1 \quad \text{(Equation 3)} \] Step 2: Solve the system of equations Start with Equation 1: \[ \frac{1}{a} - \frac{2}{c} = 1 \implies \frac{1}{a} = 1 + \frac{2}{c} \] Substitute \( \frac{1}{a} = 1 + \frac{2}{c} \) into Equation 2: \[ \left( 3 \left( 1 + \frac{2}{c} \right) \right) - \frac{1}{b} + \frac{2}{c} = 1 \] \[ 3 + \frac{6}{c} - \frac{1}{b} + \frac{2}{c} = 1 \] \[ 3 + \frac{8}{c} - \frac{1}{b} = 1 \implies \frac{8}{c} - \frac{1}{b} = -2 \quad \text{(Equation 4)} \] Now, solve Equation 3 for \( \frac{1}{b} \): \[ - \frac{3}{b} + \frac{4}{c} = 1 \implies \frac{3}{b} = \frac{4}{c} - 1 \implies \frac{1}{b} = \frac{4}{3c} - \frac{1}{3} \] Substitute this into Equation 4: \[ \frac{8}{c} - \left( \frac{4}{3c} - \frac{1}{3} \right) = -2 \] \[ \frac{8}{c} - \frac{4}{3c} + \frac{1}{3} = -2 \] Multiply through by 3 to eliminate the fraction: \[ \frac{24}{c} - \frac{4}{c} + 1 = -6 \] \[ \frac{20}{c} = -7 \implies c = -\frac{20}{7} \] Step 3: Calculate the value of \( a \) and \( b \) Substitute \( c = -\frac{20}{7} \) into Equation 1: \[ \frac{1}{a} - \frac{2}{-\frac{20}{7}} = 1 \implies \frac{1}{a} + \frac{14}{20} = 1 \implies \frac{1}{a} + \frac{7}{10} = 1 \] \[ \frac{1}{a} = 1 - \frac{7}{10} = \frac{3}{10} \implies a = \frac{10}{3} \] Substitute \( c = -\frac{20}{7} \) into Equation 3 to find \( b \): \[ - \frac{3}{b} + \frac{4}{-\frac{20}{7}} = 1 \implies - \frac{3}{b} - \frac{28}{20} = 1 \] \[ - \frac{3}{b} - \frac{7}{5} = 1 \implies - \frac{3}{b} = 1 + \frac{7}{5} = \frac{12}{5} \implies b = -\frac{15}{2} \] Step 4: Calculate \( 3a + 4b + 7c \) Now, substitute the values of \( a \), \( b \), and \( c \) into \( 3a + 4b + 7c \): \[ 3a + 4b + 7c = 3 \times \frac{10}{3} + 4 \times \left(-\frac{15}{2}\right) + 7 \times \left(-\frac{20}{7}\right) \] \[ = 10 - 30 - 20 = -15 \] Thus, the final answer is \( -15 \). \bigskip