Question

If the equation of the plane passing through the point (2, -1, 3) and perpendicular to each of the planes $3x - 2y + z = 8$ and $x + y + z = 6$ is $lx + my + nz = 1$, then $4m + 2n - 3l$ =
Identify the correct option from the following:

• A. 0
• B. $-\frac{20}{11}$
• C. 1
• D. 3

Hint:

To find the equation of a plane perpendicular to two planes, compute the cross product of their normal vectors, then use the given point to determine the constant term.

Answer 1

The Correct Option is C

Step 1: Find the normal vector of the required plane
The plane is perpendicular to the planes $3x - 2y + z = 8$ (normal: $(3, -2, 1)$) and $x + y + z = 6$ (normal: $(1, 1, 1)$). The normal vector of the required plane is the cross product of these normals: $(3, -2, 1) \times (1, 1, 1) = ((-2)(1) - (1)(1), (1)(1) - (3)(1), (3)(1) - (-2)(1)) = (-3, -2, 5)$. Thus, the equation of the plane is $-3x - 2y + 5z = d$. Step 2: Use the given point to find the equation
The plane passes through (2, -1, 3). Substitute: $-3(2) - 2(-1) + 5(3) = -6 + 2 + 15 = 11$. So, the equation is $-3x - 2y + 5z = 11$. Given the form $lx + my + nz = 1$, rewrite: $-\frac{3}{11}x - \frac{2}{11}y + \frac{5}{11}z = 1$. Thus, $l = -\frac{3}{11}$, $m = -\frac{2}{11}$, $n = \frac{5}{11}$. Step 3: Compute $4m + 2n - 3l$
$4m + 2n - 3l = 4\left(-\frac{2}{11}\right) + 2\left(\frac{5}{11}\right) - 3\left(-\frac{3}{11}\right) = -\frac{8}{11} + \frac{10}{11} + \frac{9}{11} = \frac{-8 + 10 + 9}{11} = \frac{11}{11} = 1$, which matches option (3).