Three Dimensional Geometry

Given equation of plane$\pi$is$a x+b y+11 z+d=0 \;\;\;\cdot\cdot\cdot\cdot\cdot\cdot\cdot(i)$Now, the normal vector of $\pi$ is$n_1 = a \hat{i} + b \hat{j} + 11 \hat{k} = \langle a, b, 11\rangle$Normal of first plane:$n_2 = 2 \hat{i} - 3 \hat{j} + 1 \cdot \hat{k} = \langle 2, -3, 1\rangle$Normal of second plane:$n_3 = 3 \hat{i} + \hat{j} - \hat{k} = \langle 3, 1, -1\rangle$According to question, if plane $\pi$ is perpendicular to both then its normal vector is perpendicular to the normals of the two planes.So,$n_1 \cdot n_2 = 0 \Rightarrow 2a-3b+11=0 \;\;\;\cdot\cdot\cdot\cdot\cdot\cdot\cdot(ii)$$n_1 \cdot n_3 = 0 \Rightarrow 3a+b-11=0 \;\;\;\cdot\cdot\cdot\cdot\cdot\cdot\cdot(iii)$On solving Eqs. (ii) and (iii) we get,$\;\;\; a=2 \;\text{ and }\; b=11-3a=11-6=5$$\; \therefore \langle a, b, 11\rangle = \langle 2, 5, 11\rangle$ ∴ Equation of plane will become, $2x+5y+11z+d=0 \;\;\;\cdot\cdot\cdot\cdot\cdot\cdot\cdot(iv)$ Now, distance from origin $(0,0,0)$ to the plane (iv) is$\;\;\frac{|d|}{\sqrt{2^2+5^2+11^2}} = \sqrt{6}$$\Rightarrow |d| = \sqrt{6} + \sqrt{150} = 30$$\Rightarrow d = \pm 30$(given)As question demand is all intercepts (i.e. $x, y \times z$ intercepts) are positive.So, $d$ must be negative because $x$-intercept $= \frac{-d}{2} > 0 \Rightarrow d < 0 y$-intercept$\;\; = -\frac{d}{5} > 0 \Rightarrow d < 0$$\;\; z \;\text{-intercept}\; = \frac{-d}{11} > 0 \Rightarrow d < 0$Hence, $d = -30 = -3 \times 2 \times 5 = -3 a b$