Key Concepts and Formulas

1. Coplanarity of Lines: Two lines $l_1$ and $l_2$ are coplanar if and only if there exists a plane containing both lines. If $l_2$ is given as the intersection of two planes $P_1=0$ and $P_2=0$, then any plane containing $l_2$ can be represented as $P_1 + \mu P_2 = 0$. For $l_1$ to be coplanar with $l_2$, $l_1$ must lie in this plane. A line with direction vector $\vec{d}$ and passing through point $A$ lies in a plane with normal vector $\vec{n}$ and equation $Ax'+By'+Cz'+D=0$ if:
   • $\vec{d} \cdot \vec{n} = 0$ (the line is parallel to the plane).
   • The point $A$ lies on the plane (i.e., its coordinates satisfy the plane equation).
2. Nearest Point from a Point to a Line: The point P on a line $L$ that is nearest to a given external point Q is the foot of the perpendicular from Q to $L$. This means the line segment PQ is perpendicular to the line $L$. If $\vec{d}$ is the direction vector of $L$, then $\vec{PQ} \cdot \vec{d} = 0$.

Step-by-Step Solution

Part 1: Determining the value of $\alpha$ for coplanarity

Step 1: Identify the properties of lines $l_1$ and $l_2$. The line $l_1$ is given in symmetric form: $l_{1}: \frac{x+5}{3}=\frac{y+4}{1}=\frac{z-\alpha}{-2}$ From this, we identify a point on $l_1$ as $A_1(-5, -4, \alpha)$ and its direction vector as $\vec{d_1} = (3, 1, -2)$.

The line $l_2$ is given as the intersection of two planes: $P_1: 3x+2y+z-2=0$ $P_2: x-3y+2z-13=0$

Step 2: Formulate the equation of a plane containing $l_2$. Any plane containing the line $l_2$ can be represented by a linear combination of the equations of $P_1$ and $P_2$: $(3x+2y+z-2) + \mu(x-3y+2z-13) = 0$ Rearranging this into the standard form $Ax+By+Cz+D=0$: $(3+\mu)x + (2-3\mu)y + (1+2\mu)z - (2+13\mu) = 0$ The normal vector to this plane is $\vec{n} = (3+\mu, 2-3\mu, 1+2\mu)$.

Step 3: Apply the first condition for $l_1$ to lie in the plane. For $l_1$ to lie in this plane, its direction vector $\vec{d_1}$ must be perpendicular to the plane's normal vector $\vec{n}$. Their dot product must be zero: $\vec{d_1} \cdot \vec{n} = 0$ $3(3+\mu) + 1(2-3\mu) + (-2)(1+2\mu) = 0$ $9 + 3\mu + 2 - 3\mu - 2 - 4\mu = 0$ $9 - 4\mu = 0 \implies 4\mu = 9 \implies \mu = \frac{9}{4}$ This value of $\mu$ defines the specific plane that contains $l_2$ and is parallel to $l_1$.

Step 4: Apply the second condition for $l_1$ to lie in the plane and determine $\alpha$. For $l_1$ to be coplanar with $l_2$, the line $l_1$ must actually lie in this plane. This means the point $A_1(-5, -4, \alpha)$ on $l_1$ must satisfy the plane equation. Substitute $\mu = \frac{9}{4}$ into the plane equation and then substitute the coordinates of $A_1$: $(3x+2y+z-2) + \frac{9}{4}(x-3y+2z-13) = 0$ Substitute $A_1(-5, -4, \alpha)$: $(3(-5)+2(-4)+\alpha-2) + \frac{9}{4}((-5)-3(-4)+2\alpha-13) = 0$ $(-15-8+\alpha-2) + \frac{9}{4}(-5+12+2\alpha-13) = 0$ $(\alpha-25) + \frac{9}{4}(2\alpha-6) = 0$ To eliminate the fraction, multiply the entire equation by 4: $4(\alpha-25) + 9(2\alpha-6) = 0$ $4\alpha - 100 + 18\alpha - 54 = 0$ $22\alpha - 154 = 0$ $22\alpha = 154 \implies \alpha = \frac{154}{22} \implies \alpha = 7$

Part 2: Finding the point P on $l_1$ nearest to Q

Step 5: Represent any point P on $l_1$ and the vector PQ. Now that we have $\alpha=7$, the equation of line $l_1$ is: $l_{1}: \frac{x+5}{3}=\frac{y+4}{1}=\frac{z-7}{-2}$ Let this ratio be equal to $\lambda$. Then any point $P(a,b,c)$ on $l_1$ can be expressed in terms of $\lambda$: $P(a,b,c) = (3\lambda-5, \lambda-4, -2\lambda+7)$ The given point Q is $(-4, -3, 2)$. The vector $\vec{PQ}$ is obtained by subtracting the coordinates of P from Q: $\vec{PQ} = Q - P = (-4 - (3\lambda-5), -3 - (\lambda-4), 2 - (-2\lambda+7))$ $\vec{PQ} = (-4 - 3\lambda + 5, -3 - \lambda + 4, 2 + 2\lambda - 7)$ $\vec{PQ} = (1 - 3\lambda, 1 - \lambda, 2\lambda - 5)$ Alternatively, we can use $\vec{QP} = P - Q = (3\lambda-5 - (-4), \lambda-4 - (-3), -2\lambda+7 - 2) = (3\lambda-1, \lambda-1, -2\lambda+5)$. The dot product with $\vec{d_1}$ will yield the same $\lambda$. Let's use $\vec{QP}$.

Step 6: Apply the condition for the nearest point. The line segment PQ (or QP) must be perpendicular to the line $l_1$. This means their dot product must be zero: $\vec{QP} \cdot \vec{d_1} = 0$ $(3\lambda-1)(3) + (\lambda-1)(1) + (-2\lambda+5)(-2) = 0$ $9\lambda - 3 + \lambda - 1 + 4\lambda - 10 = 0$ $14\lambda - 14 = 0$ $14\lambda = 14 \implies \lambda = 1$

Step 7: Determine the coordinates of P and calculate $|a|+|b|+|c|$. Substitute $\lambda=1$ back into the coordinates of P: $P(a,b,c) = (3(1)-5, (1)-4, -2(1)+7)$ $P(a,b,c) = (3-5, 1-4, -2+7)$ $P(a,b,c) = (-2, -3, 5)$ So, $a=-2$, $b=-3$, and $c=5$. Finally, calculate $|a|+|b|+|c|$: $|a|+|b|+|c| = |-2| + |-3| + |5| = 2 + 3 + 5 = 10$

Common Mistakes & Tips

• Incomplete Coplanarity Conditions: Remember that for a line to lie in a plane, both conditions must be satisfied: the line's direction vector must be perpendicular to the plane's normal vector, AND a point on the line must lie in the plane.
• Sign Errors: Be careful with signs when defining points on the line (e.g., $x+5$ implies $x_0=-5$) and when calculating dot products.
• Vector Subtraction: Ensure consistency when calculating the vector between two points (e.g., $Q-P$ or $P-Q$). While the dot product with the direction vector will still be zero, it's good practice to be consistent.

Summary

First, we determined the value of $\alpha$ by using the conditions for coplanarity of two lines. We formed a family of planes containing $l_2$ and found the specific plane that also contained $l_1$. This yielded $\alpha=7$. Next, we used this value of $\alpha$ to find the equation of $l_1$. We then found the point P on $l_1$ nearest to Q by ensuring the vector PQ was perpendicular to the direction vector of $l_1$. This gave us the coordinates of P as $(-2,-3,5)$. Finally, we calculated the sum of the absolute values of its coordinates, which is 10.

The final answer is $\boxed{10}$ which corresponds to option (C).