JEE Main 2026 — Mathematics Vectors & 3D Geometry

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

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

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

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

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

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

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

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

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

$$17\lambda + 1 = 0 \Rightarrow \lambda = -\dfrac{1}{17}$$17λ+1=0⇒λ=−171​

Substituting $\lambda = -\dfrac{1}{17}$λ=−171​ into the coordinates of $P$P, we get:

$$a = 2\left(-\dfrac{1}{17}\right) + 1 = \dfrac{15}{17}$$a=2(−171​)+1=1715​

$$b = -3\left(-\dfrac{1}{17}\right) + 1 = \dfrac{20}{17}$$b=−3(−171​)+1=1720​

$$c = 2\left(-\dfrac{1}{17}\right) + 1 = \dfrac{15}{17}$$c=2(−171​)+1=1715​

The sum of the coordinates is:

$$a + b + c = \dfrac{15}{17} + \dfrac{20}{17} + \dfrac{15}{17} = \dfrac{50}{17}$$a+b+c=1715​+1720​+1715​=1750​

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

$$34 \times \dfrac{50}{17} = 100$$34×1750​=100

Answer: $100$100