JEE Mathematics 2024 Question with Solution

Given: The foot of the perpendicular from $(1,2,3)$ to the line is $(\alpha, \beta, \gamma)$.

Find: The value of $19(\alpha + \beta + \gamma)$.

A point on the line is taken in parametric form as

$$x = 5t - 3, \quad y = 2t + 1, \quad z = 3t - 4$$

So the foot of the perpendicular is

$$P(\alpha, \beta, \gamma) = (5t - 3, 2t + 1, 3t - 4)$$

The vector from $A = (1,2,3)$ to $P$ is

$$\overrightarrow{AP} = (5t - 4, 2t - 1, 3t - 7)$$

Since $\overrightarrow{AP}$ is perpendicular to the line with direction ratios $(5,2,3)$, we use the dot product:

$$(5t - 4) \times 5 + (2t - 1) \times 2 + (3t - 7) \times 3 = 0$$

Expanding,

$$25t - 20 + 4t - 2 + 9t - 21 = 0$$ $$38t - 43 = 0$$

Hence,

$$t = \frac{43}{38}$$

Now substitute this value of $t$:

$$\alpha = 5t - 3 = \frac{101}{38}, \quad \beta = 2t + 1 = \frac{62}{19}, \quad \gamma = 3t - 4 = -\frac{23}{38}$$

Therefore,

$$\alpha + \beta + \gamma = \frac{101}{38} + \frac{124}{38} - \frac{23}{38} = \frac{202}{38} = \frac{101}{19}$$

Thus,

$$19(\alpha + \beta + \gamma) = 101$$

Therefore, the correct option is B.

Given: The point is $(1,2,3)$ and the line has direction ratios $(5,2,3)$.

Find: The value of $19(\alpha + \beta + \gamma)$ where $(\alpha, \beta, \gamma)$ is the foot of the perpendicular.

Write a general point on the line as

$$(x_1, y_1, z_1) = (-3 + 5t, 1 + 2t, -4 + 3t)$$

For the foot of the perpendicular, the vector joining $(1,2,3)$ to this point must be perpendicular to the line. So,

$$5(x_1 - 1) + 2(y_1 - 2) + 3(z_1 - 3) = 0$$

Substitute:

$$x_1 - 1 = -4 + 5t, \quad y_1 - 2 = -1 + 2t, \quad z_1 - 3 = -7 + 3t$$

Hence,

$$5(-4 + 5t) + 2(-1 + 2t) + 3(-7 + 3t) = 0$$ $$-20 + 25t - 2 + 4t - 21 + 9t = 0$$ $$38t - 43 = 0$$ $$t = \frac{43}{38}$$

Now,

$$\alpha = -3 + 5\cdot \frac{43}{38} = \frac{-114 + 215}{38} = \frac{101}{38}$$ $$\beta = 1 + 2\cdot \frac{43}{38} = \frac{38 + 86}{38} = \frac{124}{38}$$ $$\gamma = -4 + 3\cdot \frac{43}{38} = \frac{-152 + 129}{38} = -\frac{23}{38}$$

So,

$$\alpha + \beta + \gamma = \frac{101}{38} + \frac{124}{38} - \frac{23}{38} = \frac{202}{38} = \frac{101}{19}$$

Finally,

$$19 \times \frac{101}{19} = 101$$

Therefore, the answer is $101$, so the correct option is B.