Q40 / 75 · JEE Main 2026 (02 April Shift 1) · Mathematics

Let a line $L$ passing through the point $(1, 1, 1)$ be perpendicular to both the vectors $2\hat{i} + 2\hat{j} + \hat{k}$ and $\hat{i} + 2\hat{j} + 2\hat{k}$. If $P(a, b, c)$ is the foot of perpendicular from the origin on the line $L$, then the value of $34(a + b + c)$ is :

  • A
    $50$
  • B
    $80$
  • Correct answer
    C
    $100$
  • D
    $120$

Solution

The direction vector of the line $L$ is given by the cross product of the two vectors to which it is perpendicular:

$\vec{d} = (2\hat{i} + 2\hat{j} + \hat{k}) \times (\hat{i} + 2\hat{j} + 2\hat{k}) = 2\hat{i} - 3\hat{j} + 2\hat{k}$

The equation of the line $L$ passing through $(1, 1, 1)$ is:

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

Any point $P$ on the line $L$ has coordinates $(2\lambda + 1, -3\lambda + 1, 2\lambda + 1)$.

Since $P(a, b, c)$ is the foot of the perpendicular from the origin $(0, 0, 0)$ to the line $L$, the vector $\overrightarrow{OP}$ is perpendicular to the direction vector $\vec{d}$ of the line. Thus, their dot product is zero:

$\overrightarrow{OP} \cdot \vec{d} = 0$

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

$4\lambda + 2 + 9\lambda - 3 + 4\lambda + 2 = 0$

$17\lambda + 1 = 0 \Rightarrow \lambda = -\dfrac{1}{17}$

Substituting $\lambda = -\dfrac{1}{17}$ into the coordinates of $P$, we get:

$a = 2\left(-\dfrac{1}{17}\right) + 1 = \dfrac{15}{17}$

$b = -3\left(-\dfrac{1}{17}\right) + 1 = \dfrac{20}{17}$

$c = 2\left(-\dfrac{1}{17}\right) + 1 = \dfrac{15}{17}$

The sum of the coordinates is:

$a + b + c = \dfrac{15}{17} + \dfrac{20}{17} + \dfrac{15}{17} = \dfrac{50}{17}$

Therefore, the value of $34(a + b + c)$ is:

$34 \times \dfrac{50}{17} = 100$

Answer: $100$

Hence, the correct answer is option: 'C'.

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