Key Concepts and Formulas

• Parametric Form of a Line: A line passing through $(x_0, y_0, z_0)$ with direction ratios $(a, b, c)$ can be represented parametrically as $x = x_0 + a\lambda$, $y = y_0 + b\lambda$, $z = z_0 + c\lambda$, where $\lambda$ is a scalar parameter.
• Intersection of Two Lines: If two lines intersect, there exists a unique point that lies on both lines. This means their parametric coordinates must be equal for specific values of their respective parameters.
• Distance of a Point from a Plane: The distance of a point $(x_0, y_0, z_0)$ from a plane $Ax + By + Cz + D = 0$ is given by the formula: $D = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$

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Step-by-Step Solution

Step 1: Express Both Lines in Parametric Form

We are given two lines in symmetric (Cartesian) form. To find a general point on each line, we introduce a unique parameter for each line.

For the first line, $L_1$: $\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z + 3}{1}$ Let's set this equal to a parameter $\lambda$. $\frac{x - 1}{1} = \lambda \implies x = \lambda + 1$ $\frac{y - 2}{2} = \lambda \implies y = 2\lambda + 2$ $\frac{z + 3}{1} = \lambda \implies z = \lambda - 3$ So, any point on line $L_1$ can be represented as $P_1(\lambda) = (\lambda + 1, 2\lambda + 2, \lambda - 3)$.

For the second line, $L_2$: $\frac{x - a}{2} = \frac{y + 2}{3} = \frac{z - 3}{1}$ Let's set this equal to a different parameter, $\mu$. Using a distinct parameter is crucial because the parameters for different lines are generally independent. $\frac{x - a}{2} = \mu \implies x = 2\mu + a$ $\frac{y + 2}{3} = \mu \implies y = 3\mu - 2$ $\frac{z - 3}{1} = \mu \implies z = \mu + 3$ So, any point on line $L_2$ can be represented as $P_2(\mu) = (2\mu + a, 3\mu - 2, \mu + 3)$.

Explanation: This step converts the symmetric equations of the lines into a more flexible parametric form. This allows us to express the coordinates of any point on each line using a single variable, which is essential for finding a common point.

Step 2: Equate Coordinates to Find the Point of Intersection P

Since the lines intersect at a point P, the coordinates of $P_1(\lambda)$ and $P_2(\mu)$ must be identical for some specific values of $\lambda$ and $\mu$. We equate the corresponding $x$, $y$, and $z$ coordinates:

1. x-coordinates: $\lambda + 1 = 2\mu + a$ (Equation 1)
2. y-coordinates: $2\lambda + 2 = 3\mu - 2$ (Equation 2)
3. z-coordinates: $\lambda - 3 = \mu + 3$ (Equation 3)

Explanation: By equating the coordinates, we form a system of three linear equations. This system must have a solution for $\lambda$, $\mu$, and $a$ if the lines intersect as stated in the problem.

Step 3: Solve for Parameters $\lambda$, $\mu$, and the Unknown 'a'

We now have a system of three equations with three unknowns ($\lambda$, $\mu$, and $a$). We can solve for $\lambda$ and $\mu$ using Equations 2 and 3, which do not involve 'a'.

Let's simplify Equation 2 and Equation 3: From Equation 2: $2\lambda - 3\mu = -4$ (Simplified Eq. 2') From Equation 3: $\lambda - \mu = 6 \implies \lambda = \mu + 6$ (Simplified Eq. 3')

Now, substitute the expression for $\lambda$ from Eq. 3' into Eq. 2': $2(\mu + 6) - 3\mu = -4$ $2\mu + 12 - 3\mu = -4$ $-\mu + 12 = -4$ $-\mu = -16 \implies \mu = 16$

Now that we have $\mu = 16$, we can find $\lambda$ using Eq. 3': $\lambda = \mu + 6 = 16 + 6 = 22$

Finally, substitute the values of $\lambda = 22$ and $\mu = 16$ into Equation 1 to find $a$: $\lambda + 1 = 2\mu + a$ $22 + 1 = 2(16) + a$ $23 = 32 + a$ $a = 23 - 32$ $a = -9$

Explanation: We prioritized solving the equations that only involved $\lambda$ and $\mu$ first. This allowed us to find their specific values. Once $\lambda$ and $\mu$ were known, we substituted them into the remaining equation to determine the value of 'a', which fully defines the second line.

Step 4: Determine the Coordinates of the Point of Intersection P

Now that we have the value of $\lambda = 22$ (or $\mu = 16$), we can find the coordinates of the point of intersection P by substituting $\lambda$ into the parametric equations for $L_1$ (or $\mu$ into the parametric equations for $L_2$).

Using $\lambda = 22$ for $L_1$: $x = \lambda + 1 = 22 + 1 = 23$ $y = 2\lambda + 2 = 2(22) + 2 = 44 + 2 = 46$ $z = \lambda - 3 = 22 - 3 = 19$

So, the point of intersection P is $(23, 46, 19)$.

Explanation: With the determined parameter values, we can now find the exact coordinates of the point P where the two lines meet. It's a good practice to verify this point using the other parameter ($\mu=16$) as well: For $L_2$: $x = 2(16) + (-9) = 32 - 9 = 23$, $y = 3(16) - 2 = 48 - 2 = 46$, $z = 16 + 3 = 19$. The coordinates match, confirming our calculations.

Step 5: Calculate the Distance of Point P from the Plane $z = a$

We found $a = -9$. Therefore, the plane is $z = -9$. This can be written in the general form $Ax + By + Cz + D = 0$ as $0x + 0y + 1z + 9 = 0$. The point P is $(x_0, y_0, z_0) = (23, 46, 19)$.

Using the distance formula for a point from a plane: $D = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$ Substitute the values: $A=0, B=0, C=1, D=9$ and $(x_0, y_0, z_0) = (23, 46, 19)$. $D = \frac{|(0)(23) + (0)(46) + (1)(19) + 9|}{\sqrt{0^2 + 0^2 + 1^2}}$ $D = \frac{|0 + 0 + 19 + 9|}{\sqrt{0 + 0 + 1}}$ $D = \frac{|28|}{\sqrt{1}}$ $D = 28$

Explanation: With the value of 'a' determined, we can define the plane. The distance formula is then applied directly. For a plane like $z=k$, the distance of a point $(x_0, y_0, z_0)$ from it is simply $|z_0 - k|$. In our case, $|19 - (-9)| = |19 + 9| = |28| = 28$.

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Common Mistakes & Tips

• Using the Same Parameter: A frequent error is using the same parameter ($\lambda$) for both lines. This would incorrectly imply a relationship between the lines' parameters. Always use distinct parameters (e.g., $\lambda$ and $\mu$) for different lines.
• Algebraic Errors: Solving the system of three linear equations requires careful algebraic manipulation. Double-check substitutions and arithmetic to avoid errors.
• Understanding Plane Equation: For a plane like $z=a$, remember that $A=0, B=0, C=1$ (or $C=-1$) and $D=-a$ (or $D=a$) in the general form $Ax+By+Cz+D=0$. A shortcut for distance from $z=k$ is simply $|z_0-k|$.
• Coordinate Identification: Ensure you correctly identify $x_0, y_0, z_0$ and $A, B, C, D$ when applying the distance formula.

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Summary

We began by converting the given symmetric equations of the lines into their parametric forms using distinct parameters $\lambda$ and $\mu$. By equating the corresponding coordinates, we formed a system of three linear equations. Solving this system allowed us to find the values of $\lambda$, $\mu$, and the unknown constant 'a'. With $\lambda$ (or $\mu$), we determined the coordinates of the intersection point P. Finally, using the value of 'a' to define the plane, we calculated the perpendicular distance of point P from the plane $z=a$ using the standard formula.

The final answer is $\boxed{28}$ which corresponds to option (A).