Key Concepts and Formulas

1. Parametric Form of a Line: A line given in symmetric form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$ can be expressed in parametric form by setting it equal to a parameter, say $\lambda$. This represents any point $(x,y,z)$ on the line as $(x_1 + a\lambda, y_1 + b\lambda, z_1 + c\lambda)$.
2. Condition for a Line to Lie on a Plane: If a line lies entirely on a plane $Ax+By+Cz+D=0$, two conditions must be met:
   • The direction vector of the line $(a,b,c)$ must be perpendicular to the normal vector of the plane $(A,B,C)$. This means their dot product is zero: $Aa + Bb + Cc = 0$.
   • Any point on the line $(x_1, y_1, z_1)$ must satisfy the equation of the plane: $Ax_1 + By_1 + Cz_1 + D = 0$. Alternatively, if the parametric coordinates of a general point on the line are substituted into the plane's equation, the resulting equation must be true for all possible values of the parameter $\lambda$. This implies that the coefficients of $\lambda$ and the constant term in the resulting equation must both be zero.
3. Shortest Distance of a Plane from the Origin: The shortest distance of a plane $Ax + By + Cz + D = 0$ from the origin $(0,0,0)$ is given by the formula $d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$.

Step-by-Step Solution

Step 1: Extract information from the line and plane equations.

The equation of the line is given as: $\frac{x - 2}{3} = \frac{y + 1}{-2} = \frac{z + 3}{-1}$ From this, we can identify a point on the line $(x_1, y_1, z_1) = (2, -1, -3)$ and its direction vector $\vec{v} = (3, -2, -1)$.

The equation of the plane is given as: $px - qy + z = 5$ We can rewrite this in the standard form $Ax + By + Cz + D = 0$: $px - qy + z - 5 = 0$ From this, we can identify the normal vector of the plane $\vec{n} = (p, -q, 1)$. The constant term $D = -5$.

Step 2: Apply the condition that the line lies on the plane to find p and q.

Since the line lies entirely on the plane, two conditions must be satisfied:

• Condition 1: The direction vector of the line is perpendicular to the normal vector of the plane. The dot product of $\vec{v}$ and $\vec{n}$ must be zero: $\vec{v} \cdot \vec{n} = (3)(p) + (-2)(-q) + (-1)(1) = 0$ $3p + 2q - 1 = 0 \quad \implies \quad 3p + 2q = 1 \quad \text{(Equation 1)}$

• Condition 2: A point on the line satisfies the plane's equation. We use the point $(2, -1, -3)$ from the line and substitute it into the plane's equation $px - qy + z = 5$: $p(2) - q(-1) + (-3) = 5$ $2p + q - 3 = 5 \quad \implies \quad 2p + q = 8 \quad \text{(Equation 2)}$

Step 3: Solve the system of linear equations for p and q.

We have the system of equations:

1. $3p + 2q = 1$
2. $2p + q = 8$

From Equation 2, we can express $q$ in terms of $p$: $q = 8 - 2p$ Now substitute this expression for $q$ into Equation 1: $3p + 2(8 - 2p) = 1$ $3p + 16 - 4p = 1$ $-p + 16 = 1$ $-p = -15 \quad \implies \quad p = 15$ Substitute the value of $p$ back into the expression for $q$: $q = 8 - 2(15)$ $q = 8 - 30$ $q = -22$ So, the coefficients are $p = 15$ and $q = -22$.

Step 4: Formulate the equation of the plane.

Substitute the values of $p$ and $q$ back into the original plane equation $px - qy + z = 5$: $(15)x - (-22)y + z = 5$ $15x + 22y + z = 5$ To use the distance formula, we write it in the standard form $Ax + By + Cz + D = 0$: $15x + 22y + z - 5 = 0$ Here, $A=15$, $B=22$, $C=1$, and $D=-5$.

Step 5: Calculate the shortest distance of the plane from the origin.

Using the formula for the shortest distance $d$ from the origin $(0,0,0)$ to the plane $Ax + By + Cz + D = 0$: $d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$ Substitute the values: $d = \frac{|-5|}{\sqrt{15^2 + 22^2 + 1^2}}$ $d = \frac{5}{\sqrt{225 + 484 + 1}}$ $d = \frac{5}{\sqrt{710}}$ To simplify and match the options, we can rewrite the expression: $d = \sqrt{\frac{5^2}{710}} = \sqrt{\frac{25}{710}}$ Divide the numerator and denominator by 5: $d = \sqrt{\frac{5}{142}}$

Common Mistakes & Tips

1. Algebraic Errors: Be meticulous with signs and calculations when solving the system of equations for $p$ and $q$. A small mistake here will propagate through the entire problem.
2. Interpreting "Line on Plane": Ensure both conditions (direction vector perpendicular to normal vector, AND a point on the line satisfies the plane equation) are applied correctly. Using the parametric form and setting coefficients of $\lambda$ and constant term to zero is an equivalent and robust method.
3. Distance Formula Application: Remember the absolute value in the numerator of the distance formula, as distance is always non-negative. Also, ensure the plane equation is in the $Ax+By+Cz+D=0$ form before identifying $D$.

Summary

This problem required us to first determine the specific equation of a plane by utilizing the condition that a given line lies entirely within it. This involved setting up and solving a system of linear equations for the unknown coefficients $p$ and $q$. Once the plane's equation was established as $15x + 22y + z - 5 = 0$, we applied the standard formula to calculate its shortest distance from the origin. The calculated distance is $\sqrt{\frac{5}{142}}$.

The final answer is $\boxed{\sqrt{\frac{5}{142}}}$, which corresponds to option (B).