Question

Let P be the point of intersection of the line \(\frac{x+3}{3}=\frac{y+2}{1}=\frac{1-z}{2}\) and the plane x + y + z = 2. If the distance of the point P from the plane 3x – 4y + 12z = 32 is q, then q and 2q are the roots of the equation 

• A. \(x^2+18x+72=0\)
• B. \(x^2-18x-72=0\)
• C. \(x^2+18x-72=0\)
• D. \(x^2-18x+72=0\)

Answer 1

The Correct Option is D

The given line is represented by the equations of the form:

\[ \frac{x+3}{3} = \frac{y+2}{1} = \frac{1-z}{2} = k \]

From this, we can express \(x\), \(y\), and \(z\) in terms of parameter \(k\):

• x = 3k - 3
• y = k - 2
• z = 1 - 2k

We are given the plane equation \(x + y + z = 2\). Substitute the expressions of \(x\), \(y\), and \(z\) into this plane equation:

\[ (3k - 3) + (k - 2) + (1 - 2k) = 2 \]

Simplifying, we get:

\[ 3k - 3 + k - 2 + 1 - 2k = 2 \Rightarrow 2k - 4 = 2 \Rightarrow 2k = 6 \Rightarrow k = 3 \]

Substitute \(k=3\) back into the expressions for \(x\), \(y\), and \(z\) to get the coordinates of point \(P\):

• x = 3(3) - 3 = 6
• y = 3 - 2 = 1
• z = 1 - 2(3) = -5

Thus, the coordinates of \(P\) are \((6, 1, -5)\).

Next, find the distance of point \(P\) from the plane \(3x - 4y + 12z = 32\). The formula for the distance \(d\) from a point \((x_1, y_1, z_1)\) to the plane \(Ax + By + Cz + D = 0\) is:

\[ d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} \]

For the plane \(3x - 4y + 12z = 32\), we rewrite it as \(3x - 4y + 12z - 32 = 0\) and substitute \(P(6, 1, -5)\) into the formula:

\[ d = \frac{|3(6) - 4(1) + 12(-5) - 32|}{\sqrt{3^2 + (-4)^2 + 12^2}} \] \[ d = \frac{|18 - 4 - 60 - 32|}{\sqrt{9 + 16 + 144}} \] \[ d = \frac{|-78|}{13} = \frac{78}{13} = 6 \]

Given that this distance \(d = q\) and \(2q\), the roots of the provided equation are \(q\) and \(2q\), we have \(q = 6\) and \(2q = 12\).

To identify the correct quadratic equation whose roots are 6 and 12, consider:

\[ x^2 - (q + 2q)x + q \cdot 2q = 0 \] \[ x^2 - 18x + 72 = 0 \]

Thus, the correct answer is \(x^2-18x+72=0\).