JEE Mathematics 2023 Question with Solution

Given: $\frac{x + 3}{3} = \frac{y + 2}{1} = \frac{1 - z}{2} = \lambda$ and the plane $x + y + z = 2$.

Find: The quadratic equation whose roots are $q$ and $2q$, where $q$ is the distance of point P from the plane $3x - 4y + 12z = 32$.

From the line,

$$x = 3\lambda - 3, \quad y = \lambda - 2, \quad z = 1 - 2\lambda$$

Since point P lies on the plane $x + y + z = 2$, substitute:

$$(3\lambda - 3) + (\lambda - 2) + (1 - 2\lambda) = 2$$ $$2\lambda - 4 = 2$$ $$\lambda = 3$$

Hence,

$$P = (6, 1, -5)$$

Now the distance of $P(6,1,-5)$ from the plane $3x - 4y + 12z - 32 = 0$ is

$$q = \frac{|3(6) - 4(1) + 12(-5) - 32|}{\sqrt{3^2 + (-4)^2 + 12^2}}$$ $$q = \frac{|18 - 4 - 60 - 32|}{\sqrt{9 + 16 + 144}} = \frac{78}{13} = 6$$

So the roots are $q = 6$ and $2q = 12$.

Therefore,

$$\text{Sum of roots} = 6 + 12 = 18, \quad \text{Product of roots} = 6 \cdot 12 = 72$$

The required quadratic equation is

$$x^2 - 18x + 72 = 0$$

Therefore, the correct option is D.

Given: The line is $\frac{x + 3}{3} = \frac{y + 2}{1} = \frac{1 - z}{2}$ and the plane is $x + y + z = 2$.

Find: The equation having roots $q$ and $2q$.

1. Write the line in parameter form using a common parameter $\lambda$.
2. Find the intersection point with the plane $x + y + z = 2$.
3. Compute the perpendicular distance from that point to the plane $3x - 4y + 12z = 32$.
4. Form the quadratic using sum and product of roots.

Taking

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

we get

$$x = 3\lambda - 3, \quad y = \lambda - 2, \quad z = 1 - 2\lambda$$

Substitute these into $x + y + z = 2$:

$$3\lambda - 3 + \lambda - 2 + 1 - 2\lambda = 2$$ $$2\lambda - 4 = 2$$ $$\lambda = 3$$

Thus,

$$P = (3\cdot 3 - 3, 3 - 2, 1 - 2\cdot 3) = (6, 1, -5)$$

Now use the distance formula from point $\left(x_1,y_1,z_1\right)$ to plane $Ax + By + Cz + D = 0$:

$$\text{Distance} = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$$

Here,

$$3x - 4y + 12z - 32 = 0$$

So,

$$q = \frac{|3\cdot 6 - 4\cdot 1 + 12\cdot (-5) - 32|}{\sqrt{3^2 + (-4)^2 + 12^2}}$$ $$q = \frac{|18 - 4 - 60 - 32|}{\sqrt{9 + 16 + 144}}$$ $$q = \frac{78}{13} = 6$$

Hence the roots are $6$ and $12$. For a quadratic with roots $\alpha$ and $\beta$,

$$x^2 - (\alpha + \beta)x + \alpha\beta = 0$$

So,

$$x^2 - (6 + 12)x + (6)(12) = 0$$ $$x^2 - 18x + 72 = 0$$

Therefore, the correct option is D.