Question

If A(2,1,-1), B(6,-3,2), C(-3,12,4) are the vertices of a triangle ABC and the equation of the plane containing the triangle ABC is $53x+by+cz+d=0$, then $\frac{d}{b+c}=$

• A. -5
• B. 1
• C. 4
• D. -15

Hint:

The equation of a plane passing through three points A, B, and C can also be found using the scalar triple product in determinant form: $\begin{vmatrix} x-x_A & y-y_A & z-z_A
x_B-x_A & y_B-y_A & z_B-z_A
x_C-x_A & y_C-y_A & z_C-z_A \end{vmatrix} = 0$. This is equivalent to finding the cross product for the normal and then writing the plane equation, and can sometimes be a faster setup.

Answer 1

The Correct Option is D

Step 1: Find Normal Vector:

Vectors on the plane:
AB = (6 - 2, -3 - 1, 2 - (-1)) = (4, -4, 3)
AC = (-3 - 2, 12 - 1, 4 - (-1)) = (-5, 11, 5)

Normal n = AB × AC:

n = | i   j   k  |
    | 4  -4   3  |
    | -5 11   5  |

n = i(-20 - 33) - j(20 + 15) + k(44 - 20) = -53i - 35j + 24k

Step 2: Equation of Plane:

Equation: -53x - 35y + 24z + K = 0.
To match the given form 53x + ..., multiply by -1:
53x + 35y - 24z - K = 0

Comparing with 53x + by + cz + d = 0:
b = 35, c = -24.

Step 3: Find d:

The plane passes through A(2, 1, -1).

53(2) + 35(1) - 24(-1) + d = 0
106 + 35 + 24 + d = 0
165 + d = 0
d = -165

Step 4: Calculate Ratio:

d / (b + c) = -165 / (35 + (-24)) = -165 / 11 = -15