1. Given vectors

$\vec a=\hat i+2\hat j+3\hat k=(1,2,3)$ $\vec b=2\hat i+3\hat j-5\hat k=(2,3,-5)$ $\vec c=3\hat i-\hat j+\lambda \hat k=(3,-1,\lambda)$

2. Find $\vec b+\vec c$

$\vec b+\vec c=(2+3,\,3-1,\,-5+\lambda)=(5,2,\lambda-5)$

Since $\vec r$ is a unit vector along $\vec b+\vec c$,

$\vec r=\frac{\vec b+\vec c}{|\vec b+\vec c|}$

So,

$\vec r=\frac{(5,2,\lambda-5)}{\sqrt{5^2+2^2+(\lambda-5)^2}}$ $\vec r=\frac{(5,2,\lambda-5)}{\sqrt{29+(\lambda-5)^2}}$

3. Use the condition $\vec r\cdot \vec a=3$

Now,

$\vec r\cdot \vec a=\frac{(5,2,\lambda-5)\cdot(1,2,3)}{\sqrt{29+(\lambda-5)^2}}=3$

Compute the dot product in the numerator:

$5\cdot 1+2\cdot 2+(\lambda-5)\cdot 3=5+4+3\lambda-15=3\lambda-6$

Hence,

$\frac{3\lambda-6}{\sqrt{29+(\lambda-5)^2}}=3$

4. Solve for $\lambda$

Divide by $3$:

$\frac{\lambda-2}{\sqrt{29+(\lambda-5)^2}}=1$

Therefore,

$\lambda-2=\sqrt{29+(\lambda-5)^2}$

Since the square root is non-negative, this is valid only if $\lambda\ge 2$.

Now square both sides:

$(\lambda-2)^2=29+(\lambda-5)^2$

Expand both sides:

$\lambda^2-4\lambda+4=29+\lambda^2-10\lambda+25$

$\lambda^2-4\lambda+4=\lambda^2-10\lambda+54$

Cancel $\lambda^2$:

$-4\lambda+4=-10\lambda+54$

$6\lambda=50$

$\lambda=\frac{25}{3}$

Thus,

$3\lambda=25$

5. Check options

• A: $21$
• B: $25$ ✅
• C: $27$
• D: $30$

So the correct option is B.