JEE Mathematics 2023 Question with Solution

Given: The plane is $8x + y + 2z = 0$. The points are $A(-3, -6, 1)$ and $B(2, 4, -3)$. Point $C$ is the intersection of the plane with line $AB$. We need the direction ratios $a, b, c$ of the perpendicular from $C$ to the line $\frac{x - 1}{-1} = \frac{y + 4}{2} = \frac{z + 2}{3}$ and then find $|a+b+c|$.

Find: The value of $|a+b+c|$.

The line through $A$ and $B$ can be written as

$$\frac{x - 2}{5} = \frac{y - 4}{10} = \frac{z + 3}{-4} = \lambda$$

So a general point on $AB$ is

$$(5\lambda + 2,\; 10\lambda + 4,\; -4\lambda - 3)$$

Substitute in the plane equation:

$$8(5\lambda + 2) + (10\lambda + 4) + 2(-4\lambda - 3) = 0$$ $$40\lambda + 16 + 10\lambda + 4 - 8\lambda - 6 = 0$$ $$42\lambda + 14 = 0$$ $$\lambda = -\frac{1}{3}$$

Hence,

$$C = \left(\frac{1}{3}, \frac{2}{3}, -\frac{5}{3}\right)$$

Now consider the line

$$\frac{x - 1}{-1} = \frac{y + 4}{2} = \frac{z + 2}{3} = \mu$$

A general point on this line is

$$(-\mu + 1,\; 2\mu - 4,\; 3\mu - 2)$$

Therefore,

$$\overrightarrow{CD} = \left(-\mu + 1 - \frac{1}{3}\right)\mathbf{i} + \left(2\mu - 4 - \frac{2}{3}\right)\mathbf{j} + \left(3\mu - 2 + \frac{5}{3}\right)\mathbf{k}$$ $$\overrightarrow{CD} = \left(-\mu + \frac{2}{3}\right)\mathbf{i} + \left(2\mu - \frac{14}{3}\right)\mathbf{j} + \left(3\mu - \frac{1}{3}\right)\mathbf{k}$$

Since this is perpendicular to the given line, its dot product with the line's direction vector $(-1,2,3)$ is zero:

$$\left(-\mu + \frac{2}{3}\right)(-1) + \left(2\mu - \frac{14}{3}\right)(2) + \left(3\mu - \frac{1}{3}\right)(3) = 0$$ $$\mu - \frac{2}{3} + 4\mu - \frac{28}{3} + 9\mu - \frac{1}{3} = 0$$ $$14\mu - \frac{31}{3} = 0$$ $$\mu = \frac{11}{14}$$

Substituting this value, the direction ratios of $\overrightarrow{CD}$ are

$$(-1, -26, 17)$$

So,

$$|a+b+c| = |-1 - 26 + 17| = 10$$

Therefore, the required value is $10$.

Given: Point $C = \left(\frac{1}{3}, \frac{2}{3}, -\frac{5}{3}\right)$ lies on the plane-line intersection, and the target line has direction vector $(-1,2,3)$.

Find: The direction ratios of the perpendicular from $C$ to the line, then evaluate $|a+b+c|$.

Take a variable point on the line as

$$D(-\mu + 1,\; 2\mu - 4,\; 3\mu - 2)$$

Then

$$\overrightarrow{CD} = \left(-\mu + \frac{2}{3},\; 2\mu - \frac{14}{3},\; 3\mu - \frac{1}{3}\right)$$

For perpendicularity with direction vector $(-1,2,3)$,

$$\left(-\mu + \frac{2}{3},\; 2\mu - \frac{14}{3},\; 3\mu - \frac{1}{3}\right) \cdot (-1,2,3) = 0$$

This gives

$$\mu - \frac{2}{3} + 4\mu - \frac{28}{3} + 9\mu - \frac{1}{3} = 0$$ $$14\mu - \frac{31}{3} = 0$$ $$\mu = \frac{11}{14}$$

Now substitute:

$$-\mu + \frac{2}{3} = -\frac{11}{14} + \frac{2}{3} = -\frac{5}{42}$$ $$2\mu - \frac{14}{3} = \frac{11}{7} - \frac{14}{3} = -\frac{65}{21}$$ $$3\mu - \frac{1}{3} = \frac{33}{14} - \frac{1}{3} = \frac{85}{42}$$

So the direction ratios are proportional to

$$\left(-\frac{5}{42}, -\frac{65}{21}, \frac{85}{42}\right)$$

Multiplying by $42$,

$$(-5, -130, 85)$$

Dividing by the common factor $5$,

$$(-1, -26, 17)$$

Thus,

$$|a+b+c| = |-1-26+17| = 10$$

Therefore, the correct numerical value is $10$.