JEE Mathematics 2023 Question with Solution

Given:

• $l_1 : \frac{x + 5}{3} = \frac{y + 4}{1} = \frac{z - \alpha}{-2}$
• $l_2$ is the line of intersection of the planes

$$3x + 2y + z - 2 = 0$$

and

$$x - 3y + 2z - 13 = 0$$

• Point Q is $(-4,-3,2)$

Find: The value of $|a| + |b| + |c|$ where P$(a,b,c)$ on $l_1$ is nearest to Q.

For coplanarity, the line $l_1$ must lie in a plane from the family

$$(3x + 2y + z - 2) + \mu(x - 3y + 2z - 13) = 0$$

The direction ratios of $l_1$ are $\left(3,1,-2\right)$. Since this direction lies in the plane,

$$3(3 + \mu) + 1(2 - 3\mu) - 2(1 + 2\mu) = 0$$ $$9 - 4\mu = 0$$

So,

$$\mu = \frac{9}{4}$$

A point on $l_1$ is $(-5,-4,\alpha)$. Since this point also lies on the same plane,

$$4(-15 - 8 + \alpha - 2) + 9(-5 + 12 + 2\alpha - 13) = 0$$ $$-100 + 4\alpha - 54 + 18\alpha = 0$$

Hence,

$$\alpha = 7$$

Now parametrize $l_1$ as

$$P \equiv (3\lambda - 5, \lambda - 4, -2\lambda + 7)$$

Then the direction ratios of $PQ$ are

$$(3\lambda - 1, \lambda - 1, -2\lambda + 5)$$

Since the nearest point from Q to $l_1$ makes $PQ \perp l_1$,

$$3(3\lambda - 1) + 1(\lambda - 1) - 2(-2\lambda + 5) = 0$$

Solving,

$$\lambda = 1$$

Substituting $\lambda = 1$ into the coordinates of $P$,

$$P = (-2,-3,5)$$

Therefore,

$$|a| + |b| + |c| = 2 + 3 + 5 = 10$$

So the required value is $10$. The solution states the correct option is B, but this corresponds to option value $14$, whereas the worked solution gives $10$. Hence the defensible option from the given options is C.